Q.1. If a number is decreased by 5 and divided by 7, the result is 10. What would be the remainder if 5 is subtracted from the number and then it is divided by 8?

A. 8

B. 6

C. 7

D. 9

Sol : Option B
Let the number be x. Then (x – 5)/7 = 10 ⇒ x = 75. When 5 is subtracted from 75, we get 70.
Also when 70 is divided by 8, the remainder is 6.

Q.2. Find the two numbers such that if the first is added to 6 times the second, the sum becomes 47 and if the second is added to 6 times the first, the sum becomes 37.

A. 4, 8

B. 3, 9

C. 5, 7

D. 2, 10

Sol : Option C
Let the numbers be x and y, then x + 6y = 47 and 6x + y = 37 ⇒ x = 5 and y = 7.

Q.3. Manish got married 12 years ago. His present age is 1.5 times his age at the time of his marriage. Manish’s brother was 8 years younger to him at the time of his marriage. The age of Manish’s brother is:

A. 22 years

B. 26 years

C. 28 years

D. 30 years

Sol : Option C
Let Manish’s Present age be x years. Then his age at the time of marriage = (x – 12) years. Therefore x =1.5(x - 12) ⇒ x = 36. Manish’s brother’s age at the time of his marriage = (x – 12) – 8 = (x - 20) = 16 years. Therefore Manish’s brother’s present age = (16 + 12) years = 28 years.

Q.4. W is as much younger than X as he is older than Y. If the sum of the ages of X and Y is 60 years, what is the difference between X and W’s age?

A. 3 years

B. 4 years

C. 5 years

D. Data Insufficient

Sol : Option D
X – W = W – Y ⇒ 2W = X + Y. Also, X + Y = 60 ⇒ 2W = 60. So W = 30. But since the value of X cannot be calculated, so we cannot find the difference between the ages of X and W.

Q.5. An employer engaged a servant with free boarding and lodging for one year with the condition that the servant will be given Rs. 15000 and a cycle at the end of the year. The servant agreed but served the employer only for 9 months and thus received Rs. 11000 and a cycle. The price of the cycle is:

A. Rs. 1000

B. Rs. 1250

C. Rs. 800

D. Rs. 900

Sol : Option A
12 months wages = Rs. 15000 + cycle; and 9 months wages = Rs. 11000 + cycle ⇒ 3 months wages = Rs. 4000, so 12 month wages = Rs. 16000. Therefore price of the cycle = 16000 – 15000 = Rs. 1000.

Q.6. The sum of four numbers is 144. If you add 5 to the first number, 5 is subtracted from the second number, the third is multiplied by 5 and the fourth is divided by 5, then all the results are equal. What is the difference between the largest and the smallest of the original numbers?

A. 54

B. 96

C. 68

D. Data Insufficient

Sol : Option B
Let the four numbers be A, B, C and D. Let A + 5 = B – 5 = 5C = D/5 = x.
Then, A = x – 5, B = x + 5, C = x/5 and D = 5x.
A + B + C + D = 144 ⇒ (x – 5) + (x + 5) + x/5 + 5x = 144 ⇒ x = 20.
Thus, the numbers are 15, 25, 4 and 100. Therefore required difference = (100 – 4) = 96.

Q.7. A boy has enough money to buy 40 pens. If each pen costs 50 paise less, he could buy four more pens and still have 160 paise left. How much money had he originally?

A. Rs. 240

B. Rs. 208

C. Rs. 204

D. Rs. 216

Sol : Option C
Let the original price of each pen = Rs. x. Therefore 40x = 44(x – 0.5) + 1.60 ⇒ x = Rs. 5.1. Therefore he had 40 × 5.1 = Rs. 204

Q8. If the length of a certain rectangle is decreased by 8 cm and the breadth is increased by 6 cm, then we get a square with the same area as the original rectangle. The perimeter of the original rectangle (in cm) is:

A. 100

B. 80

C. 90

D. 120

Sol : Option A
Let the length and breadth of the rectangle be x and y cm, respectively.
Then, (x – 8) (y + 6) = xy or 6x – 8y = 48 ........................... (i)
Also, (x – 8) = (y + 6) [Sides of Square] or x – y = 14........... (ii)
From (i) and (ii), x = 32 and y = 18.
Perimeter of the original rectangle = 2(x + y) = 100 cm

Q9. In a certain get together, there was a basket of fruits for every three guests, a basket of cookies for every four of them and a basket of chips for every seven of them. If in all, there were 122 baskets of eatables, then how many guests were there in the get together?

A. 128

B. 142

C. 168

D. 156

Sol : Option C
Let the number of fruit baskets be x, the number of cookies baskets bowls be y and the number of chips baskets be z.
Now, x + y + z = 122 ....... (i)
3x = 4y = 7z ..................................... (ii)
From (i) and (ii), we have x = 56, y = 42, z = 24.
Thus, total number of guests = 3x = 4y = 7z = 168.

Q10. There are two conference halls A and B. If 20 delegates are sent from hall A to hall B, the number of delegates in each hall is the same, while if 40 are sent from hall B to hall A, the number of delegates in hall A becomes triple the number in hall B. The number of delegates in each hall is, respectively:

A. 140 and 100

B. 100 and 80

C. 80 and 20

D. 100 and 60

Sol : Option A
According to question, we have A - 20 = B + 20 ⇒ A – B = 40............................... (i)
And 3(B – 40) = (A + 40) ⇒ 3B – 120 = A + 40 or A – 3B = - 160........................... (ii)
From (i) and (ii), we get A = 140 and B = 100