Latest Teaching jobs   »   Maths Questions for DSSSB Exam :...

Maths Questions for DSSSB Exam : 23rd March 2018 (Solutions)

Q1. A started a work and left after working for 3 days. Then B was called and he finished the work in 13 1/2 days. Had a left the work after working for 4 1/2 days, B would have finished the remaining work in 9 days. In how many days can each of them, working alone, finish the whole work?
(a) 7.5 days, 22.5 days
(b) 7 days, 9 days
(c) 5 days, 15 days
(d) 23.5 days, 8.5 days
Q2. Some amount out of Rs 9000 was lent at 8% per annum and the remaining at 6% per annum. If the total simple interest from both the fractions in 7 years was Rs 3900, find the sum lent at 8% per annum.
(a) Rs 5000/7
(b) Rs 6000/7
(c) Rs 1600
(d) Rs 1700
Q3. An article is sold at 22 1/2% profit. If its cost price is increased by Rs 40 and at the same time if its selling price is also increased by Rs 35, the percentage of profit becomes 15%. Find the cost price.
(a) Rs 156 1/3
(b) Rs 156 2/3
(c) Rs 146 1/3
(d) Rs 146 2/3
Q4. 8 years ago, the ratio of the ages of Rachana and Archana was 4 : 3. If the ratio of their present ages is 6 : 5 respectively, what is the ratio of the sum to the difference of their present ages?
(a) 11 : 2
(b) 11 : 4
(c) 11 : 1
(d) 11 : 3
Q5. In how many different ways the letters of word ‘COMBINATION’ can be arranged in order to keep the consonants always together ?
(a) 58500
(b) 48000
(c) 64800
(d) 64400
Q6.A shopkeeper calculated his profit on the marked price and finds it to be 30%. He forgets the fact that he gave a discount of 20%. What is his actual profit percentage ?
(a) 100/7%
(b) 20%
(c) 25%
(d) 200/7%
Q7.In the year 2002, the cost price of an item was 90% of the selling price while in 2003, the cost price was 95% of the selling price. If the gross profit remains the same for both the years, what is the percentage increase in the cost price of the item from 2002 to 2003 ?
(a) 120%
(b) 95%
(c) 111%
(d) 105%
Q8.Mayank can do 50% more work than Shishu in the same time. Shishu alone can do a piece of work in 30 hours. Shishu starts working and he had already worked for 12 hours when Mayank joins him. How many hours should Shishu and Mayank work together to complete the remaining work ?
(a) 6
(b) 12
(c) 4.8
(d) 7.2
Q9.There are 12 pipes connected to a tank. Some of them are fill pipes and the others are drain pipes. Each of the fill pipes can fill the tank in 8 h and each of the drain pipes can drain completely in 6 h. If all the pipes are kept open, an empty tank gets fill in 24 h. How many of the 12 pipes are fills pipes ?
(a) 5
(b) 6
(c) 7
(d) 8
Q10.A car goes 20 miles on a gallon of gas driving at 60 miles/h. If the car is driven at 70 miles/h it goes as far as 70%. How many gallons of gas will be required to travel 210 miles/hr.
(a) 6.17
(b) 22.5
(c) 12
(d) 15
Solutions:
S1. Ans.(a)
Sol.
3/A+27/2B=1           ……….(i)
=9/2A+9/B=1       ………(ii)
Multiply   (i) by 3/2  and on solving
A will take 7.5 days and B will take 22.5 days.
S3. Ans.(d)
Sol.
Let the CP be P then SP is 1.225 P
Now CP = P + 40, SP = 1.225P+35
15=(1.225P+35-P-40)/(P+40)×100
15=(0.225P-5)/(P+40)×100
15P+600=22.5P-500
1100=7.5P
P=146  2/3
S4. Ans.(c)
Sol.
(Rachana’s  present age)/(Archana’s present age)=6x/5x
8 years age the ratio of their age =4/3=(6x-8)/(5x-8)
20x-32=18x-24
2x=8
x=4
Rachana’s age = 24 years
Archana’s age = 20 years
Required ratio = 44/4=11/1
S5. Ans.(c)
Sol.
ATQ, CMBNTN are to be kept together and O, I, are repeated twice
∴ Total no. of different ways =(6!×6!)/(2!×2!×2!)=64,800
S6. Ans.(a)
Sol.
When profit is calculated on M.P.
then C.P. = 100 – 30 (let M.P. = Rs. 100)
= 70
and S.P. = Rs. 80
∴Actual profit=(80 –70)/70×100
=14 2/7%
S7. Ans.(c)
Sol.
Let S.P. in year 2002 was Rs. x.
And in year 2003 was Rs. y
In 2002, C.P. = 0.9x
In 2003, C.P. = 0.95y
Since profit in both years is same
∴ x – 0.9x = y – 0.95y
⇒ 2x = y
∴ Required percentage
=(0.95×2 –0.9)/0.9×100
= 111.11
≃ 111%
S8. Ans.(d)
Sol.
Mayank can do the whole work alone in
30×2/3=20 hours
Let in x hours they complete the remaining work together
∴(12+x)/30+x/20=1
⇒ 24 + 2x + 3x = 60
⇒ x = 7.2 hours
S9. Ans.(c)
Sol.
Let x are fill pipes out of 12 pipes
∴ x fill pipes can fill the tank in 8/x hours
And (12 – x) drain pipes can drain the tank in 6/(12–x) hours
∴  x/8–(12–x)/6=1/24
3x – 48 + 4x = 1
⇒ 7x = 49
⇒ x = 7
S10. Ans.(d)
Sol.
In one gallon gas, car goes 20 miles at the speed 60 miles/hr.
In one gallon gas, car will go
=20×70/100=14 miles
At a speed of 70 miles/hr.
Now, if car has to cover 210 miles in one hours,
Required number of gallons of gas
=210/14=15 gallons