Q1. When the numerator of a fraction is increased by 4, the fraction increases by 2/3. What is the denominator of the fraction?
(a) 6
(b) 7
(c) 8
(d) None of these
Sol.
Let the fraction be x/y
According to condition
(x+4)/y=x/y+2/3
⇒x/y+4/y=x/y+2/3
⇒4/y=2/3⇒y=6
Q2. A pencil costs two and a half rupees. Amit buys one and a half dozen pencils and gives a 100 rupee note to the shopkeeper. The money he will get back is
(a) Rs. 30
(b) Rs. 55
(c) Rs. 45
(d) Rs. 65
Sol.
Cost of 1 pencil = Rs. 2.50
Amit bought one and a half dozen
= 12 + 6 = 18 pencils
The money he will get back
=100-(18×2.50)=100-45.00=Rs.55
Q3. A shop has 239 toys. Seventy more toys were brought in then 152 of them were sold. The number of toys left was
(a) 239 + 70 – 152
(b) 239 – 70 – 152
(c) 239 + 70 + 152
(d) 239 – 70 + 152
Sol.
Total number of toys = 239
Since, seventy more toys were brought in
∴ Total number of toys = 239 + 70
Now, 152 toys were sold
∴ Left toys = 239 + 70 – 152
Q4. The difference of 5671 and the number obtained on reversing its digits is
(a) 7436
(b) 3906
(c) 4906
(d) 3916
Sol.
Given number = 5671
After reversing its digits = 1765
∴ Difference = 5671 – 1765 = 3906
Q5. In the following, which is the greatest number?
(a) [(2+2)^2 ]^2
(b) (2+2+2)^2
(c) 〖(4)〗^2
(d) 〖(2×2×2)〗^2
Sol.
(4)^2=16
〖(2×2×2)〗^2=8^2=64
[(2+2)^2 ]^2=[〖(4)〗^2 ]^2=〖16〗^2=256
〖(2+2+2)〗^2=6^2=36
∴ Required number = [(2+2)^2 ]^2
Q6. A chocolate has 12 equal pieces. Manju gave one-fourth of it to Anju, one-third of it to Sujatha and one-sixth of it to Fiza. The number of pieces of chocolate left with Manju is
(a) 4
(b) 1
(c) 2
(d) 3
Sol.
Anju had =1/4×12=3 pieces
Sujatha had = 1/3×12=4 pieces
Fiza had = 1/6×12=2 pieces
Total pieces = 3 + 4 + 2 = 9 pieces
Choclate left with Manju = 12 – 9 = 3 pieces
Q7. What should be added to the product 140×101 to get 14414?
(a) 364
(b) 264
(c) 274
(d) 278
Sol.
Product of 140×101=14140
Difference = 14414-14140=274
274 is added to the product to get 14414.
Q8. What should be added to the product 103×301 to get 31103?
(a) 301
(b) 103
(c) 110
(d) 100
Sol.
Product of 103×301=31003
Now, 100 is to be added to get 31103.
Q9. Number of common factors of 12 and 16 is ____.
(a) 2
(b) 3
(c) 4
(d) 5
Sol.
Factors of 12 are = 2×2×3
Factors of 16 are = 2×2×2×2
Common factors of 12 and 16 are 2×2
Number of common factors of 12 and 16 are 2.
Q10. In a hockey match between School A and School B, School A scored 11 goals and School B scored 3 goals. What fraction of the total goals did School A score?
(a) 3/14
(b) 8/11
(c) 11/14
(d) 11/3
Sol.
Total goals = School A’s goals + School B’s goals = 11 + 3 = 14
