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# Maths Quiz for 2016-17 Exams

Q1. The average of 100 numbers is 44. The average of these 100 numbers and four other, new numbers is 50. The average of the four new numbers will be
(a) 800
(b) 200
(c) 176
(d) 24
Sol.
Sum of 100 numbers = (44 × 100) = 4400
Sum of 104 numbers (50 × 104) = 5200
Sum of 4 numbers (5200 – 4400) = 800
Average of these 4 numbers = 800/4 = 200

Q2. The average of marks of 28 students in mathematics was 50; 8 students left the school and then the average increased by 5. What is the average of marks obtained by the students who left the school?
(a) 37.5
(b) 42.5
(c) 45
(d) 50.5
Sol.
Total marks of 28 students = (28 × 50) = 1400
Average marks of 20 students = (50 + 5) = 55
Total marks of 20 students = (55 × 20) = 1100
Average marks of these 8 students = 300/8 = 37.5

Q3. Six persons went to a hotel for taking their meals. Five of them spent Rs. 32 each on their meals while the 6th person spent Rs. 80 more than the average expenditure of all the six. Total money spent by all the persons is:
(a) Rs. 192
(b) Rs. 240
(c) Rs. 288
(d) Rs. 336
Sol.
Let the average expenditure of all the six be Rs. x.
Total expenditure = Rs. 6x
(32 × 5) + (x + 80) = 6x
⇒ 5x = (160 + 80) = 240
⇒ x = 48
Total expenditure = Rs. (6 × 48) = Rs. 288

Q4. There were 35 students in a hostel. If the number of students be increased by 7, the expenditure on food increases by Rs. 42 per day while the average expenditure of students is reduced by reduced by Rs. 1. What was the initial expenditure of food per day?
(a) Rs. 432
(b) Rs. 442
(c) Rs. 420
(d) Rs. 400
Sol.
Let the original expenditure be Rs. x
Then, x/35-((x+42))/42=1
⇒ 6x – 5(x + 42) = 210
⇒ x = 420
∴ Original expenditure = Rs. 420

Q5. In a college, 40% of the students were allotted group A, 75% of the remaining were given group B and the remaining 12 students were given group C. Then the number of students who applied for the group is
(a) 100
(b) 60
(c) 80
(d) 92
Sol.
Group A = 40%
Group B = (60×75)/100= 45%
Group C = 15%
If the total number of students be x, then
=(x×15)/100=12
⇒x=(12×100)/15=80

Q6. A tradesman gives 4% discount on the market price and gives 1 article free for buying every 15 articles and thus gains 35%. The marked price is increased above the cost price by:
(a) 20%
(b) 39%
(c) 40%
(d) 50%
Sol.
Let the C.P. of each article be Rs. x
Then, C.P. of 16 articles = Rs. 16x
S.P. of 15 articles = 135% of Rs. 16x
=Rs.(135/100×16x)=Rs.108x/5
S.P. of 1 article = Rs. (108x/5×1/15) = Rs. 36x/25
If S.P. is Rs. 96, M.P. = Rs. 100
If S.P. is Rs. =36x/25, M.P. = Rs. (100/96×36x/25)
= Rs. 3x/2
Increase in M.P. over C.P.
=((3x/2-x)/x)×100
= 50%

Q7. A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole, the quantity sold at 18% profit is
(a) 500 kg
(b) 600 kg
(c) 400 kg
(d) 640 kg
Sol.
Let the sugar sold at 8% gain = x
∴ Sugar sold at 18% gain = (1000 – x)
Let CP of sugar = Rs. y per kg
Total CP = Rs. 1000y
∴(108/100×xy)+118/100 (1000-x)y
= 114/100 × 1000y
⇒ 108xy + 118000y – 118xy = 114000y
⇒ 10x = 4000
∴ x = 400
∴ Quantity sold at 18% profit
= (1000 – 400) kg = 600 kg

Q8. From a container of wine, a thief has stolen 15 litres of wine and replaced it with same quantity of water. He again repeated the same process. Thus in three attempts the ratio of wine and water became 343:169. The initial amount of wine in the container was:
(a) 150 litres.
(b) 120 litres.
(c) 135 litres.
(d) 125 litres.
Sol.
It means = (wine (left))/(wine (initian amount))=343/512
(∴ 343 + 169 = 512)
Thus, 343x = 512x(1-15/k)^3 ⇒343/512=(7/8)^3=(1-15/k)^3
⇒(1-15/k)=(7/8)
⇒ k = 120
Thus the initial amount of wine was 120 litres.

Q9. A student walks from his house at 5/2 km/h and reaches his school late by 6 min. Next day, he increase his speed by 1 km/h and reaches 6 min before school time. How far is the school from his house ?
(a) 5/4 km
(b) 7/4 km
(c) 9/4 km
(d) 11/4 km
Sol.
According to the formula,
Required distance = ab(t1+t2 )/(b – a)
Where a = 5/2=5/2, b = 7/2, t1 = 6, t2 = 6
∴ Required distance = (5/2  × 7/2 ((6 + 6)/60))/(7/2  – 5/2)
=35/4×12/60=7/4  km

Q10. A car run 1 km in 4 min 50 sec and B in 5 minutes. How many metres’ start can A give B in a km race so that the race may end in a dead heat?
(a) 33.33m.
(b) 25m.
(c) 20m.
(d) 40m.
Sol.
Time taken by A to run 1 km = 290 seconds.
Time taken by B to run 1 km = 300 seconds.
A can give B a start of (300 – 290) sec, i.e. 10 sec.
In 300 sec, B runs 1000 m.
In 10 sec, B runs (1000/300×10) m = 331/3 m
∴ A can give B a start of 100/3 m.