Q1. Sum of cost price of two watches is Rs 3360. One watch is sold at a profit of 12% and other at a loss of 12%. Due to this transaction, there is neither profit nor loss. Find the cost price of each watch.
(a) Rs 1680, Rs 1680
(b) Rs 1280, Rs 1280
(c) Rs 1000, Rs 1000
(d) Rs 2100, Rs 2160
Sol.
On one watch, he earns 12% profit and on the other one 12% loss. There is, finally, no profit no loss.
Hence CPs of both the watches should be same such that their summation = Rs 3360. Hence CP/watch = (Rs.3360)/2 = Rs. 1680
Hence option (a) is the answer.
Q2. A seller buys 50 kg oranges for Rs 525 and gains a profit equal to the cost price of 10 kg of oranges. Find the selling price per kg of oranges (Rs/kg).
(a) 12.60
(b) 11.60
(c) 10.20
(d) 12
(e) 13.125
Sol.
Assume SP/kg = Rs x/kg.
CP/kg =(Rs.525)/50 = Rs. 10.5/kg
Total SP – Total CP = profit ⇒ 50x – Rs. 525 = 10 × Rs. 10.5 = Rs. 105
Or, 50x = Rs. 105 + Rs. 525 = Rs. 630
Hence x =(Rs.630)/50 = Rs. 12.6/kg
Q3.Find a number, one-seventh of which exceeds its eleventh part by 100.
(a) 1925
(b) 1825
(c) 1540
(d) 1340
Sol.
Let the number be x.
x/7-x/11 = 100
=>(11x-7x)/(11×7)=100
=> 4x = 77 x 100
=> x = 1925
Q4.A can do a work in 12 days. When he had worked for 3 days, B joined him. If they complete the work in 3 more days, in how many days can B alone finish the work?
(a) 6 days
(b) 12 days
(c) 4 days
(d) 8 days
Sol.
Let be alone do the work in x days.
6 x 1/12+3×1/x =1
=>1/2+3/x =1
=>3/x= 1/2=>x = 6 days
Q5. Mohan invested a certain sum in a simple interest bond whose value grew to Rs. 300 at the end of 3 yr and to Rs. 400 at the end of another 5 yr. The rate of interest was
(a) 12%
(b) 12.5%
(c) 8.33%
(d) 6.67%
Sol.
(Pr*3)/100 +P = 300…….(1)
(Pr*8)/100 +P = 400…….(2)
subtract (1) from (2)
p*r = 2000 ………..(3)
put (3) in (1)
p = 240, then r = 8.33%
Q6. Krishna’s father was five times as old as Krishna 5 years ago. After 10 years, he will be three times as old as Krishna. What is the average age of Krishna and his father today?
(a) 10 years
(b) 50 years
(c) 30 years
(d) 25 years
Sol.
Let Krishna’s present age be x years and his father’s present age be y years.
Now, 5(x – 5) = (y – 5) ⇒ 5x – y = 20 …(i)
Again, 3(x + 10) = y + 10 ⇒ 3x – y = -20 …(ii)
From equation (i) and (ii) we get
x = 20, y = 80
So, present average age = 50 years
Q7. The average weight of 6 women decreases by 3 kg when one of them weighting 80 kg is replaced by a new woman. The weight of the new woman is
(a) 62 kg
(b) 60 kg
(c) 70 kg
(d) 72 kg
Sol.
Weight of new woman = weight of replaced woman – Reduced average weight of 6 woman
= 80 – 6 × 3
= 62 kg
Q8. How many litres of fresh water should be mixed with 30 litres of 50% milk solution so that resultant solution is a 10% milk solution?
(a) 120 litres
(b) 25 litres
(c) 150 litres
(d) 60 litres
Sol.
In initial milk solution
Milk = 15 litres
Water = 15 litres
According to the question
15/(30+x)=1/10=milk/Total
150 = 30 + x
x = 120 litres
Q9. A person bought 20 kg of rice at Rs. 10.50 per kg and 30 kgs at Rs. 12 per kg. He mixed the two and sold the mixture at Rs. 11.40 per kg. What was the gain in this deal (in Rs)?
(a) 0
(b) 20
(c) 100
(d) 75
Sol.
Total cost of 50 per kg = 20 × 10.5 + 30 × 12
= 210 + 360 = 570
Total sell cost = 50 × 11.4 = 570
Gain = Total sell – Total cost
= 570 – 570 = 0
Q10. In an examination, Hari’s average marks was 63 per paper. If he had obtained 20 more marks for his geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination?
(a) 8
(b) 9
(c) 10
(d) 11
Sol.
Let total paper were in examination = x
Total marks he got = x×63
According to question,
x×63+20+2=65x
2x = 22
x = 11
