7. | The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is: | |||||||
Answer: Option A Explanation:
Let the numbers be a, b and c. Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400. (a + b + c) = 400 = 20. |
8. | A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by: | |||||||
Answer: Option D Explanation:
Let the ten’s digit be x and unit’s digit be y. Then, number = 10x + y. Number obtained by interchanging the digits = 10y + x. (10x + y) + (10y + x) = 11(x + y), which is divisible by 11. |
9. | In a two-digit, if it is known that its unit’s digit exceeds its ten’s digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is: | |||||||
Answer: Option A Explanation:
Let the ten’s digit be x. Then, unit’s digit = x + 2. Number = 10x + (x + 2) = 11x + 2. Sum of digits = x + (x + 2) = 2x + 2. (11x + 2)(2x + 2) = 144 22x2 + 26x – 140 = 0 11x2 + 13x – 70 = 0 (x – 2)(11x + 35) = 0 x = 2. Hence, required number = 11x + 2 = 24. |
10. | Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number. | ||||||||||
Answer: Option A Explanation:
Let the number be x.
x2 + 17x – 60 = 0 (x + 20)(x – 3) = 0 x = 3. |