CAT Quantitative Aptitude Questions for practice
CAT Quantitative Aptitude Questions for practice
Directions for questions 1 to 5 : Refer to the data below and answer the questions that follow.
I. The ratio of people doing all 3 activities to people doing atleast 2 activities is 1 : 6.
II. The ratio of people doing only one activity to people doing atleast 2 activities is 3 : 2
III. Number of people doing only CAT/MBA exceeds number of people doing GRE/MS only by 990.
IV. Number of people doing only GRE/MS exceeds number of people seeking jobs only by 360.
1. Number of people doing all 3 activities is :
1] 500
2] 1000
3] 1500
4] Cannot be determined
2. Number of people doing no more than on activity is :
1] 10,000
2] 9,760
3] 14,000
4] 12,000
3. Number of people doing exactly two activities is :
1] 6000
2] 5000
3] 9000
4] 10,000
4. The number of people not doing any activity is :
1] 760
2] 570
3] 1000
4] Data Insufficient
5. The number of people who are both seeking employment and trying for GRN/MS is :
1] 1610
2] 2430
3] 1720
4] Data Insufficient
Explanatory Answers :
For answers to questions 1 to 5 :
Let the different sections be represented by the variables.
Given
1. p : (x + y + z + p) = 1 : 6
2. (a + b + c) : (x + y + z + p) = 3 : 2
∴ (a + b + c) : p : (x + y + z + p) = 9 : 1 : 6
Since P = 1
x + y + z + p 6
Since P = 1
x + y + z 5
∴ (a + b + c) : (x + y + z) : p = 9 : 5 : 1 —— (I)
Again, the total of all areas is :
(a + b + c) + 2 (x + y + z) + 3p = 6810 + 7070 + 8120
(a + b + c) + 2 (x + y + z) + 3p = 22000
Using (I)
9p + 2(5p) + 3p = 22p = 22000
∴ p = 1000
x + y + z = 5000 } (II)
a + b + c = 9000
Again,
c – b = 990
b – a = 360
On solving
a = 2430
b = 2790
c = 3780
p = people doing all 3 activities = 1000
a + b + c = people doing exactly 1 activity = 9000
x + y + z = people doing exactly 2 activities = 5000
people not doing any activity = EU – (1000 + 9000 + 5000)
= 15760 – 15000 = 760
x, y, z cannot be determined without additional data.
Hence,
1 [2] 2[2] 3[2] 4[1] 5[4]
Directions for questions 6 to 9 : Refer to the data below and answer the questions that follow.
Palaash never knew he would be in such trouble. He has to sing in seven different languages. The Assamese song would be a hit if and only if it is preceded by a Telugu song. Konkani song would be a hit only if a Hindi song came before it. English has to be preceded by Hindi and Punjabi is the last song he sings. There is one song between English and Punjabi and of the three only two were hit. Konkani song is a failure. Tamil is the seventh language. If the Assamese song is a hit and his first song was Telugu then:
6. Which song did he sing after English?
1] Telugu
2] Konkani
3] Tamil
4] Assamese
7. Which song did he sing after Konkani?
1] Tamil
2] Punjabi
3] Hindi
4] English
8. If each hit had two points in the Grammy awards, then how many points did Palaash get?
1] 4
2] 6
3] 8
4] Can’t say
9. What were the total number of assured hit songs as per the question?
1] 3
2] 4
3] 5
4] 6
Explanatory Answers :
For answers to questions 6 to 9 :
From the clues we can see Telugu is the first song, while Punjabi is the last song. Assamese would be second followed by Konkani, Hindi and English. Thus Tamil, the seventh language would be the sixth song. In terms of points there is no information regarding whether Telugu and Hindi songs were a hit or not. Thus we can’t say. But regarding the number of assured hits as per the question it would be Assamese, and two songs out of English, Tamil and Punjabi. Thus the answers are
6 [3] 7[3] 8[4] 9[1]
Directions for questions 10 to 14 : Refer to the data below and answer the questions that follow.
A six digit (with no digit repeated) number is such that every alternate digit is a prime number. The difference between the digit on the tens place and the digit on the thousands place is the digit on the lakh place. The units digit is the product of the digit on the lakh place and the digit on the tenthousands place, which is also a prime number. Also tens place digit is greater than the thousandth place digit.
10. The digit on the lakh place is :
1] 1
2] 2
3] 3
4] 4
11. The digit on the units place is :
1] 2
2] 4
3] 6
4] 8
12. The positive difference of the the digit on the tenthousandth place and the the ten’s place is :
1] 3
2] 4
3] 6
4] 7
13. The digit on the hundred’s place is :
1] 3
2] 4
3] 8
4] Can’t say
14. The number is divisible by :
1] 2
2] 3
3] 4
4] Both [1] and [3]
Explanatory Answers :
For answers to questions 10 to 14 :
As the unit digit is not a prime number, thus the lakh digit, the thousands digit and the ten’s digit will be a prime number.
As the lakh digit is the difference of two prime numbers and still is a prime number, it is 2. Also the digits on the thousandth and the ten’s place can be either 3 and 5 or 5 and 7. As the unit digit is the product of the digit on the lakh place and the digit on the tenthousandth 7 (as 5 x 2 = 10, 7 x 2 = 14). Thus it has to be 3. Therefore we see that the digit on the thousandth place is 5 and that on the ten’s place is 7. We don’t know anything about the digit on the hundredth place. Therefore the number is 235 x 76
Thus the answer are :
10 [2] 11[3] 12[2] 13[4] 14[4]
Directions for questions 15 to 19 : Refer to the data below and answer the questions that follow.
In a gathering of 8 people, A, B, C and D are Software professionals and P, Q, R and S are Hardware professionals. Each person shows atleast one of the following features viz. HK, JM, GS and CP.
HW : Hardware professional SW : Software Professional
HK : Has Kids JM : Just Married
GS : Has grand sons CP : One of the Couples
 A ‘CP’ may be JM (and vice versa) but definitely not HK or GS.
 A GS can not be JM.
 Neither JM nor CP can be HK or GS.
 GS implies HK but the reverse need not be true.
 P, Q, R, D and S hav only one feature where as others show atleast 1.
 A is JM, where as one of C and D is both CP and JM.

B is GS but D and A are not HK.

D and Q always show the same feature.

R is JM and S is HK.

The 4 features are shown by atleast one of the SW or HW.

No professional can show more than two features.
15. How many professionals are definitely not HK?
1] 2
2] 3
3] 5
4] 5 or 6
16. A does not show all of the features except :
1] JM
2] GS
3] Data insufficient
4] None of these
17. If A does not show CP then Q does show :
1] HK
2] JM
3] GS
4] Data insufficient
18. P can show all of the following except :
1] JM
2] GS
3] CP
4] None of these
19. If P is the only HW who is a CP then which of the following is definitely true?
I. A does not show CP
II. B may show CP
III. D show JM
1] I only
2] II only
3] III only
4] None of these
Explanatory Answers :
For answers to questions 15 to 19 :
The following matrix can be formed based on the given information
SW  HW  
A  B  C  D  P  Q  R  S  
HK  ×  ü  ×  ×  ×  ü  
JM  ü  ×  ü  ü  ×  
GS  ×  ü  ×  ×  ×  ×  ×  × 
CP  ×  ü  ×  × 
15. Five professionals are definitely not HK, P may or may not be HK. Hence, [3].
16. Now A may or may not be CP. Hence [3].
17. We don’t know whether Q is JM or CP. Hence, [4].
18. P cannot show GS. Hence, [2].
19. I : A may or may not show CP. Hence, I may be false.
II : B cannot show more than 2 features. Hence, II is false.
III : D should show JM as Q will show JM. Hence, III is true. Hence, [3].
Directions for questions 20 to 23 : Refer to the data below and answer the questions that follow.
I have got two uncles, Joe and John and two aunts, Jane and Jill. While uncle Jose is not married, uncle John is married and has two sons as well. Aunt Jane is a widow and has only one daughter. My mother is the only one in the family who has a real sister while she has no brothers. My father works in the diamond factory while my paternal grandparents live with us. We are a joint family. Then :
20. How many members are there in our family?
1] 10
2] 11
3] 12
4] 13
21. How many cousin sisters do I have?
1] None
2] One
3] Two
4] Three
22. Who is married to Uncle John?
1] Jane
2] Jill
3] None of these
4] Can’t Say
23. How many children do my paternal grandparents have?
1] Two
2] Three
3] Four
4] Five
Explanatory Answers :
For answers to questions 20 to 23 :
The family heirarchy will be like this :
20. 12 members in the family. Hence, [3].
21. Only one cousin sister. Hence, [2].
22. Aunt Jill is married to Uncle John. Hence, [2].
23. 3 sons, my father, Uncle Joe and Uncle John. Hence, [2].
24. The total number of female members in the family is 5. Hence, [4].
25. The years can be calculated as:
1980 – 15 + 1 + 10 + 2 + 6 = 1984. Hence, [3]
Directions for questions 24 to 25 : Choose the correct alternative.
24. Mr. Gupta is Rita and Sita’s fatherinlaw. Amit is married to Sita and has got two daughters. Rita has only one son. If Mr. Gupta has only two sons and Mrs. Gupta is alive then how many female members are there in the Gupta faimly?
1] 2
2] 3
3] 4
4] 5
25. I build a time machine in 1980. The peculiar ;thing about the time machine was that it had a range of twenty years into the future and 30 years into the past. I used the machine and went back 15 years. After spending a year in the past, I gain used it and went 10 years into the future. There I spend around two years and finally went to the future by 6 years. In which year did I land up?
1] 1980
2] 1983
3] 1984
4] None of these
Directions for questions 23 to 30 : Refer to the data below and answer the questions that follow.
Cars and motorcycles continued to surge ahead even as “Made in India” automobiles like commercial vehicles, multiutility vehicles, scooters, mopeds and threewheelers failed to attract overseas buyers this fiscal. Car exports shot up by a huge 105% at 42,444 units during 200102 on the back of a superb performance by Ford India, which accounted for more than half of the total exports. Ford shipped 27,572 units of its midsize car “Ikon” to countries like South Africa and Mexico. Car market leader Maruti Udyog, however, posted a 35.7% dip at 8,559 units during 200102. South Korea’s Hyundai also witnessed a 14.5% decline in exports at 4,494 units (5250 cars last year). Car exports of Telco surged ahead by 261.6% at 1,689 cars (467 cars last year) while that of General Motors slipped to 40 cars yearonyear from 69 units. In sharp contrast to passenger car segment, commercial vehicle exports slumped by 19.9% at 9,683 units. While medium and heavy (M&H) vehicles exports went down by 18.2% to 3,891 units, that of light commercial vehicles (LCVs) declined by 22.3% at 5,792 units. Telco’s M&H exports fell by 24.4% to 2,118 units while LCVs dipped by 36.2% to 4,036 vehicles during 200102. Ashok Leyland, India’s second largest commercial vehicle marker, reported a 6.7% land 4.8% drop in M&H and LCV exports at 1,725 units and 98 units respectively. However, exports of LCV markers like Eicher, Swaraj Mazda and Mahindra & Mahindra rose by 61.6%, 56.5% and 68.2% at 1,062, 321 and 274 units respectively. Twowheeler exports went down by 9.2% to 91,731 units as scooter and moped exports declined by 15.7% and 41.7% at 19,369 units and 22,801 unit5s respectively. Howeer, exports of motorcycles recorded a 13% rise at 17,490 units while tha tof Bajaj Auto rose by 158% to 15,951 units. Hero Honda posted a 29.4% rise at 11,833 units. Exports of TVS Motors dipped by 13.7% to 2,273 units while Royal Enfield saw a rise of 63.6% to 1,357 units. Scooter markers like Bajaj and TVS posted a 31.5% and 137% jump in exports at 7,996 and 897 units respectively while LML and Kinetic exports dipped by 27.8% and 57.5% to 6,709 and 3,702 units.
26. If only 5 companies, Ford, Maruti Udyog, Hyundai, Telco and General Motors exported cars in 200001, how many cars did Ford export in 200001?
1] 1645
2] 1607
3] 1700
4] 1555
27. By how much were the total vehicle exports by Telco in 200102, higher or lower than those in 200001? (Telco manufactures only cars and commercial vehicles)
1] 18.3% lower
2] 15% lower
3] 12% higher
4] almost same
28. What were the total vehicle exports in 200001?
1] 133940
2] 135000
3] 132750
4] 131680
29. If the average export realization of motorcycles has reduced from Rs.45,000 per unit in 200001 to Rs.40,000 per unit in 200102, what is the change in export volumes?
1] Rs.218 million increase
2] Rs.218 crore increase
3] Rs.2.18 billion increase
4] Rs.2180 million increase
30. If the average export realization per car for Maruti Udyog has increased by 40%, then what is the percentage change in car exports for Maruti in value terms?
1] 10% increase
2] 10% decrease
3] 5% decrease
4] 5% Increase
Explanatory Answers :
For answers to questions 20 to 23 :
Car  200102  200001  M&H  200102  200001 
Ford  27572  1607  Telco  2118  2802 
Maruti  8559  13311  Ashok Ley.  1725  1849 
Hyundai  4494  5250  
Telco  1689  467  3891  4757  
GM  40  69  
90  
42444  20704 
LCV  200102  200001  Motorcycles  200102  200001 
Scooters 
200102  200001 
Telco  4036  63265  Yamaha  17490  Bajaj  7996  
Ashok Ley.  98  103  TVS  15951  TVS  897  
Eicher  1062  657  Royal Enfield  11833  LML  6709  
Swaraj  321  205  Kinetic  3702  
M&M  274  163  49561  39200  
19369  22976  
5792  7454  Mopeds  22801  39110 
42444
26. Total car exports in 200001 = (1 + 1.05) = 20704
Car exports by Maruti in 200001 = 8559 = 8559/0.643 = 13311
(1 – 0.357)
Car exports by Hyundai in 200001 = 5250
Car exports by Telco in 200001 = 467
Car exports by General Motors in 200001 = 69
Therefore, car exports by Ford in 200001 = 20704 – 13311 – 5250 – 467 – 69 = 1607 units. Hence, [2].
27. Number of vehicles exported by Telco in 200102 = 1689 + 2118 + 4036 = 7843
Car exported by Telco in 200001 = 467
Multiutility vehicles exported by Telco in 200001 = 2118 = approx. 2118/0.756 = 2802
(10.244)
LCVs exported by Telco in 200001 = 4036 = approximately 4036 = 6326
(10.362) 0.64
Total vehicle exports by Telco in 200101 = 467 + 2802 + 6326 = 9595
Therefore, ratio of Telco’s exports in 200102 to those in 200001 = 7843 = approximately 7845 = 0.817 = 81.7%
9595 9600
Therefore, Telco’s exports in 200102 were 18.3% lower. Hence, [1].
28. Car exports in 200001 = 20704
M&H vehicle exports in 20001 = 3891 ≈ 3900/0.82 = 4756
(10.182)
LCV exports in 200001 = 5792 = 7454
(10.223)
Twowheeler exports in 200001 = 91731 = 101025
(10.092)
Therefore, total vehicle exports in 200001 = 20704 + 4757 + 7454 + 101025 = 133939. Hence, [1].
29. Motorcycle export value in 200102 = 49561 × 4000 = 1982440000 or Rs.1982.44 million
Motorcycle exports in 200001 = 49561 ≈ 49000/1.25 = 39,200
1.272
Motorcycle exports value in 200001 = 39200 × 45000 = 1764000000 or Rs.1764 million
Therefore, motorcycle exports have increased by (19821764) = Rs.218 million. Hence [1].
30. Let x units be exported by Maruti Udyog for an average realization of y rupees per car in 200001. Therefore, exports of Maruti Udyog in 200001 = x × y rupees
Exports of Maruti Udyog in 200102 = x × (10.357) × y × 1.40 = 0.90 × x × y Therefore, exports of Maruti Udyog have decreased in value terms by about 10%. Hence,[2].
Directions for questions 31 to 35: Refer to the data below and answer the questions that follow.
Table shows Births to date and Lifetime births expected per 1,000 women for Selected Years from 1976 to 1998 (Numbers in thousands and age of women at survey date.)
Source : U.S. Census Bureau
Internet Release date : October 18, 2001
Characteristics  Total, 18 to 34 years  18 to 24 years  25 to 29 years  30 to 34 years  
Births to date  Lifetime births expected  Births to date  Lifetime births expected  Births to date  Lifetime births expected  Births to date  Lifetime births expected  
1998  1,104  2,045  532  1,936  1,150  2,082  1,662  2,127 
1992  1,135  2,098  521  2,053  1,181  2,137  1,679  2,106 
1990  1,130  2,116  537  2,062  1,152  2,152  1,695  2,135 
1988  1,095  2,073  459  2,045  1,163  2,116  1,674  2,057 
1987  1,125  2,074  503  2,057  1,219  2,111  1,689  2,055 
1986  1,114  2,099  497  2,087  1,171  2,117  1,734  2,094 
1985  1,098  2,062  508  2,046  1,193  2,113  1,674  2,029 
1983  1,096  2,079  481  2,071  1,220  2,082  1,786  2,088 
1982  1,086  2,023  453  1,994  1,241  2,026  1,792  2,059 
1981  1,136  2,048  517  2,033  1,273  2,012  1,857  2,106 
1980  1,127  2,059  507  2,023  1,238  2,022  1,905  2,150 
1979  1,144  2,072  514  2,033  1,269  2,033  1,942  2,170 
1978  1,171  2,113  496  2,033  1,324  2,060  2,066  2,297 
1977  1,197  2,133  485  2,052  1,345  2,049  2,150  2,351 
1976  1,263  2,160  528  2,030  1,442  2,098  2,266  2,445 
Note : Only women reporting on number and future births expected are included in this table.
Source : Current Population Survey, June 2000.
31. What year has the maximum ratio of births to date to life time birth expected in the category of 18 to 24 ?
1] 0.25
2] 0.27 (Ans)
3] 0.29
4] 0.31
Ans : For the year 1998 the ratio is highest = 532/1936 = 0.27.
32. Which year has the maximum ratio birth to date to lifetime birth expected in 3034 category?
1] 1981
2] 1978
3] 1977
4] 1976 (Ans)
Ans : For the year 1976 the ratio is 0.93, which is higher than ratios of other years.
33. In the year 1999, if the birth to date is expected to rise by 10% while lifetime birth expected is to remain same in the category of 2529. What is the new ratio of the birth to date to lifetime birth expected in year 1999 in the category of 2529 ?
1] 0.5
2] 0.6 (Ans)
3] 0.7
4] None of these
Ans : New ratio = 1150 × 1.1 = 0.61 ≈ 0.6
2028
34. In which year lifetime birth expected has shown maximum percentage increase in the category of 1824 as compared to previous year?
1] 197778
2] 197879
3] 198283 (Ans)
4] 198890
Ans : 198283 the percentage increase is by 4% which is more than other alternatives.
35. What is the maximum decrease in the births to date from 19761998 in any of the categories?
1] 101
2] 107
3] 116
4] None of these (Ans)
Ans : In the year 197879 there is decrease in births to date by 124 in 30 to 34 category.
Directions for questions 36 to 40: Refer to the data below and answer the questions that follow.
The table below shows establishments by group of total area, by Mesoregions, Microregions and Municipalities in Brazil.
Mesoregions, Microregions and Municipalities  Establishments by groups of total area (ha), on 12.31.1995  
Less than 10  10 to less than 100  100 to less than 200  200 to less than 500  500 to less than 2000  2000 and over  Not informed  
Norte 
134 803  217 097  52 061  23 477  12 333  3 799  2 605 
Rondonia  17 618  43 581  10 591  3 389  1 398  377  21 
Acre  3 962  13 647  3 753  1 528  742  156  105 
Amazonas  43 793  34 066  3 237  1 314  482  130  267 
Roraima  1 025  2 990  1 771  760  504  345  81 
Para  64 838  104 435  24 180  7 955  3 478  1 313  205 
Amapa  953  1 095  739  297  140  51  74 
Tocantins  2 614  17 283  7 790  8 234  5 589  1 427  1 976 
Nordeste  1 570 511  604 261  67 596  43 996  19 504  3 217  17 328 
Maranhao  272 100  59 360  11 207  7 267  3 370  633  14 254 
Piaui  134 949  55 192  8 888  5 250  2 274  445  1 113 
Ceara  245 312  76 199  9 472  5 711  2 259  264  385 
36. What is the percentage difference between Norte and Maranhao in the less than 10 category?
1] 98%
2] 100%
3] 102% (Ans)
4] 104%
Ans : Percentage difference = (272100 – 134803) × 100 = 102%
134803
37. What is the sum of establishments in 100 to less than 200 category?
1] 218135
2] 201185 (Ans)
3] 205035
4] 215685
Ans : After summation we get value of 201285 ≈ 201000.
38. By how much the sum of establishments of Maranhao is greater than that of Piaui?
1] 140000 ha
2] 150000 ha
3] 160000 ha (Ans)
4] 170000 ha
Ans : Difference = 368191 – 208111 = 160080 ≈ 160000.
39. Sum of which area is the greatest?
1] Maranhao
2] Piaui
3] Ceara
4] Norte (Ans)
Ans : Norte has the highest value of 446175.
40. Find the lowest total area amongst the various groups. (less than 10 etc.)
1] Roraima
2] Acre
3] Amapa
4] Rondonia (Ans)
Ans : Rondonia has the lowest value of 21 in not informed category.
Directions for questions 31 to 35: Refer to the data below and answer the questions that follow.
Give below is the data of M/s. Siddhi Chemicals from 19951999 (in Rs.’000)
I41. What is the profit after tax for the year 1998?
1] Rs. 1.3 lakhs
2] Rs. 0.065 lakhs
3] Rs. 0.65 lakhs (Ans)
4] None of these
Ans : Revenue for 1998 = Rs.600000
Cost for 1998 = Rs. 500000
Profit after tax = 0.65(600000 – 500000) = 65000 = 0.65 lakhs.
42. In which year was the percentage change in profit after tax, minimum with respect to previous years percent change in profit after tax?
1] 1996 (Ans)
2] 1997
3] 1998
4] 1999
Ans :
199596 200 → 100 (50%)
199697 100 → 300 (+200%)
199798 300 → 100 (66.67%)
199899 100 → (100) (+200%)
∴ The % change is minimum in 1996.
Note : % change in profit after tax and % change in profit before tax shall be same as tax rate is constant throughout.
43. What is the total tax collection from 19951998?
1] Rs. 2.45 lakhs (Ans)
2] Rs. 02.1 lakhs
3] Rs. 24,0500
4] Rs. 21,000
Ans : Total profit before tax = Rs.700,000
(19951998)
Tax = 35% of Rs.700000 = Rs.245000.
44. If Margin = Profit after tax × 100, what is the average margin from 19951999 ?
Revenue
1] 18%
2] 14% (Ans)
3] 21%
4] None of these
Ans : Margin 1995 = 130/600 ≅ 22%
1996 = 65/400 ≅ 16%
1997 = 195/500 ≅ 39%
1998 = 65/600 ≅ 11%
1999 = 100/500 ≅ 20% (assuming no tax benefit)
∴ Average Margin = 22 + 16 + 39 + 11 – 20/5 = 68/5 = 13.6% ≅ 14%
45. What shall be the profit before tax in year 2000, if the annual growth rate of revenue and cost is same as in 1999?
1] 2.5 lakhs
2] 3 lakhs (Ans)
3] 2 lakhs
4] 3.5 lakhs
Ans : Revenue in 2000 = 500000 × 5/6 ≅ Rs.417000
Cost in 2000 = 600000 × 1.2 = Rs.720000
Profit before tax = 417000 – 720000 ≅ – Rs.3 lakhs (approximately).
Directions for questions 46 to 55 : Each of the following questions is followed by two statements.
Mark [1] , if the question can be answered by using any of the statements alone but not by using the other statement alone.
Mark [2] , if the question can be answered by using either of the statements alone.
Mark [3] , if the question can be answered only by using both the statements together.
Mark [4] , if the question cannot be answered.
46. Decide whether the square root of the integer x is an integer.
I. The last digit of x is 2.
II. x is divisible by 3.
Ans : The key fact to solving this problem is that digit of a square of an integer is the same as the last digit of the integer squared. For example, the last digit of the square of 94 is 6, which is the same as the last digit of the square of 4, which is 16. If you square each digit {0, 1, 2,…., 8, 9}, you will see that the only possible last digits for a square are 0, 1, 4, 5, 6 and 9. Thus, if the last digit of x is 2, x can not be a square. So the square root of x is not an integer. So statement I alone is sufficient. Since 12 is divisible by 3 and is not a square but 36, which is also divisible by 3, is a square, statement II alone is not sufficient. Hence, [1]
47. How tall is the tallest among the three people?
I. The average height of three persons is 170 cms.
II. The shortest person is 145 cms.
Ans : Average height can not give any idea of tallest person and height of shortest person does not give any solution alone. Thus, each statement alone is insufficient. Combining I and II:
A + B + 145 = 170 x 3
One equation and two variables, data is insufficient. Hence, [4].
48. Is N an odd number?
I. 2N is an even number
II. 7N is an even number
Ans : From I : 2 N is even : Any number multiply by 2gives even number, so N can not be identified.
Thus, I alone is not sufficient.
From II : 7N is an even number, only if N is an even number. Therefore N is an even number. Hence, [1].
49. What is the value of two digit number whose digits are x and y?
I. The difference of x and y is 8.
II. If the digits are reversed, the number increased by 27.
Ans : First statement gives x – y = 8.
No single solution can satisfy second statement 10y + x – 10x – y = 27
9x – 9y = 27
x – y = 3
No single solution satisfy
Both the statements contradict each other. Hence, [4].
50. What is the value of the positive integer K?
I. K is divisible by 16 and 24
II. The square of K is less than 84,100
Ans : Each statement alone cannot give any single solution. Square of K less than 84100 means, k less than 290 LCM of 16, 24 is 48
Number less than 290 and divisible by 48 is 96, 144, 192, 240 and 288.
Therefore, unique solution for k can be found. Hence, [4]
51. How many students failed in both English and Math?
I. 20 passed in Math and 15 passed in English
II. The percentage of students who passed in Math was 80
Ans : Each statement alone is not sufficient to answer.
Total number of students was 25 (80% students passed in Maths = 20)
It give no idea of how many students failed in both the subjects. Hence, [4].
52. The three numbers a, b, c are all positive and are in geometric progression. Is the common ratio of the GP less than 1?
I. a, b and c are all fractions less than 1.
II. 1/a is less than 1/b.
Ans : Statement I doesn’t give any information
Statement II gives a > b, i.e. b/a < 1
Therefore a > b > c, it means common ratio is less than 1. Thus, II alone is sufficient. Hence, [1].
53. a, b, c are integers and a/(bc) = 3. What is the value of (bc)/b?
I. a/b = 3/7.
II. a and b have no common factor greater than 1.
Ans : From statement I a = 3b/7
a = 3 x (b – c)
∴ 3 x (b – c) = 3b/7
(b – c)/b = 1/7
Statement 2 gives no relation between a and b. Hence, [1].
54. The owners of a plant need to send a shipment to a distributor in a different town. They can ship either by truck or by ship. If the trucking fee is based on miles and the ship is based on tons, which is the less expensive means of transportation for this shipment?
I. The location of the distributor is 500 km away.
II. The trucking charges are Rs.500 plus Rs.2 per km, and the water freight charges are Rs.1500 plus Rs.4,000 per ton.
Ans : The truck fee is based on miles and the water freight fee is based on tons. So you will need to find something like the price per Km for the trucking and how far the trucks have to go and the price per ton for ship freight and how many tons need to be transported.
Statement I : This statement tells you nothing about the cost of sending the shipment by truck or by ship freight, so it is insufficient.
Statement II : This tells you nothing about the distance between the cities, which you would need to know in order to find the cost of sending the shipment by truck. Further, it does not tell you the weight of the shipment, which you would need to know in order to find the cost of sending the shipment by water freight.
Statements I and II : You still can’t find the cost of sending the shipment by ship freight because you don’t know the weight of the shipment. Hence, [4].
55. The shard price of Wipro has risen consistently for 10 years. What would be its worth in December 1998?
I. In December 1994, the share was valued at Rs.50.
II. During the bull market from 1994 to 1999, the share value increased by 10% each year.
Ans : Statement I : Alone is obviously insufficient.
Statement II : You need to know what the stock was worth at some time between 1994 and 1999. So, II alone is insufficient.
By using I and II together you can figure out the worth of a share in December 1998. Hence, [3].
Directions for questions 56 to 61: Refer to the data below and answer the questions that follow.
The terms of GP’s are related to the terms of an AP in the following way :
 The first term and common difference of the AP are a and d respectively.
 The number of GP’s is equal to the number of terms in the AP.
 The first term of all the GP’s is same as the first term of the AP.
 The common ratio of nth GP is ‘a’ less than the nth term of the AP.
 The number of terms in any GP is equal to that in the AP.
 The first GP’s has ‘unity’ common ratio.
 ‘a’ is a positive natural number.
56. If the common difference of the AP is 1/2, what is the largest term amongst AP and GP’s given that AP consists of 9 terms?
1] a + 4
2] 4a
3] (4)^{8}a (Ans)
4] (9/2)^{a}a
Ans :
AP a a+d a+2d a+3d …….. a + 8d ……..
GP_{1} a a a a …….. a …….
GP_{2} a ad ad^{2} ad^{3} …….. ad^{8} ……..
GP_{3} a a(2d) a(2d)^{2} a(2d)^{3} …….. a(2d)^{8} ……..
GP_{4} a a(3d) a(3d)^{2} a(3d)^{3} …….. a(3d)^{8} ……..
– – – – – – – –
– – – – – – – –
Since AP has 9 terms, its last term = a + 8d = a + 4.
Second GP here is GP_{2} with highest term a.
Since ‘a’ is positive and natural and d is a fraction any other term of a GP is less than a.
GP_{3} has common ratio = 1/2 x 2 = 1.
Hence, GP_{3} here, is same as GP_{1}
GP_{4} has common ratio = 3/2. Its highest term = a (3/2)^{8}
and so on.
The largest tem will be for GP_{9} which is a (8/2)^{8} = a(4)^{8}. Hence, [3].
57. If the common difference of the AP is an Egyptian fraction, at least how many GP’s will be exactly identical? (An Egyptian fraction is one in which numerator is ‘1’ and denominator is a positive integer more than 1).
1] All the GP’s may be unique
2] 2
3] More than 2 but finite number
4] Infinitely many
Ans : As seen in previous question if d = 1/2; GP_{1} and GP_{3} are the same
If d = 1/3 ⇒ GP_{1} and GP_{4 }will be identical and so on Hence, the minimum number of identical GP’s is 2. Note, that [1] is definitely false. Hence, [2].
58. In previous question what is the maximum number of identical GPs if d is not an Egyptian fraction?
1] All the GP’s may be unique
2] 2
3] More than 2 but finite number
4] Infinitely many
Ans : Consider a case when d is 1, all the GP’s will be same. The number of GP’s may be infinite as the number of terms in the AP may be infinite. Hence, [4].
59. What is the total number of terms (including the terms of AP and all GPs) if the number of terms of AP is a perfect square i.e., x^{2 }?
1] x^{2} + x
2] (x^{2} + x)^{2}
3] (x^{2} + x)^{2} – 2(x)^{3}
4] x^{4} – x^{2}
Ans : If number of terms in AP = x^{2}
Total number of GP_{s} = x^{2}, each with x^{2 } number of terms
Hence, total number of elements in all = x^{2 }+ x^{4 }
Consider [3]. (x^{2 }+ x)^{2} – 2x^{3} = x^{4 }+ x^{2 }+ 2x^{3} – 2x^{3}
= x^{4 } + x^{2}
Hence, [3].
60. If the second term of the second GP and the 3rd term of the third GP are 6 and 144 respectively, then what is the sum of first 10 terms of an AP ?
1] 105
2] 280
3] 310
4] Data insufficient
Ans : If a and d are the first term and common difference respectively
GP_{2} (2) = ad = 6 — (i)
GP_{3} (3) = a(2d)^{2 } = 144 — (ii)
Divide (ii) ÷ (i)
∴ 4d = 144/6 = 24 ⇒ d = 6
Put d = 6 in (i) ⇒ a = 1
Hence, ∑ (AP) = n/2 (2a + (n – 1)d) = 10/2 (2(1) + 9 x 6) = 5 (2 + 54) = 280
n=10
Hence, [2].
61. In the above question, what is the sum of all the terms in the series (inclusive of AP and GPs)?
_{ 9}
1] 290 + ∑ [(6n)^{10} – 1/5!]
^{ n=1}
_{10}
2] 280 + ∑ [(6n)^{10} — 1/(6n – 1)]
^{ n=1}
_{ 9}
3] 310 + ∑ [(6n)^{10} – 1/5n]
^{ n=0}
_{ 9}
4] 290 + ∑ [(6n)^{10} – 1/(6n – 1)]
^{ n=1}
Ans : Since a = 1, d = 6 and n = 10
Hence, S_{total } = S_{AP } + ∑ S_{GPn}
^{ n=10}
S_{GP1} = a + a + …. a = 10
S_{GP2} = a . d^{10} – 1/d – 1 = 1 6^{10} – 1/5
S_{GP3} = a . (2d)^{10} – 1/(2d – 1) = (12)^{10} – 1/11
S_{GP4} = a (3d)^{10} – 1/(3d – 1) = 18^{10} – 1/17
———
S_{GP10} = a (ad)^{10} – 1/(ad – 1) = 54^{10} – 1/53
_{ }
Now, S (total) = 280 + ∑ S_{GPn}
^{ n=10}
_{ 9}
280 + 10 + ∑ [(6n)^{10} – 1/(6n – 1)]
^{ n=0}
Which is same as [4] . Hence, [4].