Practice Quantitative aptitude - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) A shopkeepers sells an article at 12$1/2$% loss. If he sells it for Rs. 92.50 more, then he gains 6%. What is the cost price of the article?
(a)
(b)
(c)
(d)
S.P. = 100 – 12.5 = Rs. 87.5
S.P. after 6% gain = Rs. 106
Difference = Rs. 18.5
∴ C.P. = $92.5/18.5 × 100$ = Rs.500
Q-2) The mean of 20 observations is 17. On checking it was found that the two observations were wrongly copied as 3 and 6. If wrong observations are replaced by correct values 8 and 9, then what is the correct mean ?
(a)
(b)
(c)
(d)
Mean of 20 observation = 17
total sum = 17 × 20 = 340
ATQ ${340 - 3 - 6 + 8 + 9}/20 = 348/20$ = 17.4
Q-3) A container contained 80 kg of milk. From this container, 8 kg of milk was taken out and replaced by water. This process was further repeated two times. How much milk is now contained by the container?
(a)
(b)
(c)
(d)
(e)
Amount of liquid left after n operations, when the container
originally contains x units of liquid from which y units in
taken out each time is x $({x - y}/x)^n$ units.
Thus, in the above case, amount of milk left
= 80$[{80 - 8}/80]^3$ kg = 58.32 kg
Q-4) A’s income is 60% of B’s income, and A’s expenditure is 70% of B’s expenditure. If A’s income is 75% of B’s expenditure, find the ratio of A’s savings to B’s savings.
(a)
(b)
(c)
(d)
Let B's Income = Rs. x
A's Income = Rs. $3/5x$
And B's expenditure = Rs. y
A's expenditure = Rs. $7/10y$
Also, $3/5x = 3/4 . 7/10 y$
$\text"A' savings"/\text"B' savings" = {x - y}/{3/5x - 7/10y}$
= ${7/8y - y}/{3/5 . 7/8 y - 7/10y} = {- y/8}/{{21y}/40 - 7/10 y} = 5/25$ ≈ 1 : 5
Q-5) $1/2$ of $3/4$ of a number is 2$1/2$ of 10. What is the number?
(a)
(b)
(c)
(d)
Let the number is x.
According to the question
$1/2$ of $3/4$ of x = 2$1/2$ of 10
=$\text"3x"/8 = 5/2$ × 10
x = ${5×10×8}/{3×2} = 200/3 =66{2/3}$
Q-6) The solution of linear inequalities x + y ≥ 5 and x – y ≤ 3 lies
(a)
(b)
(c)
(d)
x+ y ≥ 5
let first draw graph of x + y = 5 ... (i)
put y = 0 in (i)
x = 5
put x = 0 in (i)
y = 5
Checking for (0, 0)
x + y ≥ 5
0 ≥ 5, which is false
Hence, origin does not lie in x + y ≥ 5
Now,
x – y ≤ 3
let first drawn graph of x – y = 3 ...(ii)
put x = 0 in (ii)
y = –3
put y = 0 in (ii)
x = 3'
Checking for (0, 0)
x – y ≤ 3
0 ≤ 3, which is true
Hence, origin lies in x – y ≤ 3
So, shaded region will be the solution of these linear inequalities lies in the first and second quadrant.
Q-7) If A = {x : x is an even natural number},
B = {x : x is a natural number and multiple of 5} and
C = {x : x is a natural number and multiple of 10},
then what is the value of A ∩ (B ∪ C)?
(a)
(b)
(c)
(d)
We know that
A ∩ (B ∪ C) = {A ∩ B) ∪ (A ∩ C)
Example:
A = Set of an even natural number
A = {2, 4, 6, 8, 10, 12, ...}
B = Set of natural number and multiples of 5.
B = {5, 10, 15, 20, 25, ...}
C = Set of natural number and multiple of 10.
C = {10, 20, 30, 40, 50, ...}
A ∩ B = {2, 4, 6, 8, 10, 12, ...} ∩ {5, 10, 15, 20, 25, ...} = {10, 20, 30, ...}
A ∩ C = {2, 4, 6, 8, 10, 12, ...} ∩ {10, 20, 30, 40, 50, ...} = {10, 20, 30, 40, ...}
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
= {10, 20, 30, ...} ∪ {10, 20, 30, 40, ...}
= {10, 20, 30, 40, ...}
Q-8) The ratio of two numbers is 3 : 4 and their HCF is 5. Their LCM is :
(a)
(b)
(c)
(d)
If the numbers be 3x and 4x,
then
HCF = x = 5
∴ Numbers = 15 and 20
∴ LCM = 60
Q-9) The population of a town increases every year by 4%. If its present population is 50,000, then after 2 years it will be
(a)
(b)
(c)
(d)
Using Rule 17,
Required population = $50000(1 + 4/100)^2$
= $50000 × 26/25 × 26/25$= 54080
Q-10) A shopkeeper purchased a TV for Rs.2,000 and a radio for Rs.750. He sells the TV at a profit of 20% and the radio at a loss of 5%. The total loss or gain is
(a)
(b)
(c)
(d)
S.P. of TV = $2000 × 120/100$ = Rs.2400
S.P. of radio = ${750 × 95}/100$ = Rs.712.5
Total S.P. = 2400 + 712.5 = Rs.3112.50
Gain = 3112.5 - 2000 - 750 = Rs.362.50
Q-11) If X= {a, {b}, c}, Y = {{a}, b, c} and Z = {a, b, {c}}, then (X ∩ Y) ∩ Z equals to
(a)
(b)
(c)
(d)
(X ∩ Y) = {a,{b},c} ∩ {{a},b,c} = c
Now, (X ∩ Y) ∩ Z
= c ∩ {a, b,{c}} = Φ
Q-12) Consider the following statements:- The product of any three consecutive integers is divisible by 6.
- Any integer can be expressed in one of the three forms 3k, 3k + 1, 3k + 2, where k is an integer.
Which of the above statements is/are correct ?
(a)
(b)
(c)
(d)
I. The product of any three consecutive integers is divisible by 3! i.e., 6.
II. Here, 3k = {..., 6, 3,0,3,6,... - - }
3k + 1 = {..., 5, 2,1,4,7,... - - }
and 3k + 2 = {..., 4, 1, 2,5,8,... - - }
∴ ++ {3 ,3 1,3 2 kk k }
= ------ {..., 6, 5, 4, 3, 2, 1,0,1, 2,3, 4,5,6,...}
Hence, it is true.
Q-13) A sells an article to B at a gain of 10%, B sells it to C at a gain of 5%. If C pays Rs.462 for it, what did it cost to A ?
(a)
(b)
(c)
(d)
Let the C.P. for A be Rs.x, then
$x × 110/100 × 105/100 = 462$
$x = {462 × 100 × 100}/{110 × 105}$ = Rs.400
Using Rule 15,
Here, $r_1 = 10%, r_2$ = 5%
C.P. for C = C.P. for A
$(1 + r_1/100)(1 + r_2/100)$
462 = C.P. for A
$(1 + 10/100)(1 + 5/100)$
C.P. for A = ${462 × 100 × 100}/{110 × 105}$ = Rs.400
Q-14) The average marks of 65 students in a class was calculated as 150. It was later realised that the marks of one of the students was calculated as 142, whereas his actual marks were 152. What is the actual average marks of the group of 65 students ? (Rounded off to two digits after decimal)
(a)
(b)
(c)
(d)
(e)
Actual average marks
= ${65 × 150 + 152 - 142}/65$
= ${9750 + 10}/65$ = 150.15
Q-15) The average weight of first 11 persons among 12 persons is 95 kg. The weight of 12th person is 33 kg more than the average weight of all the 12 persons. The weight of the 12th person is
(a)
(b)
(c)
(d)
Weight of 12th person = x kg (let).
∴ Average weight of 12 persons
= $({11×95+x}/12)$kg
According to the question,
${11×95+x}/12$ +33 = x
⇒ 1045 + x + 396 = 12x
⇒ 1441 = 11x
⇒ x = $1441/11$ = 131 kg.
Q-16) Average weight of 19 men is 74 kgs, and the average weight of 38 women is 63 kgs. What is the average weight (rounded off to the nearest integer) of all the men and the women together?
(a)
(b)
(c)
(d)
(e)
Avg. wt. of man & women
= ${19 × 74 + 38 × 63}/{19 + 38}$
= $3800/51$ = 66.67 ~ 67.
Q-17) If $27^{2x - 1} = (243)^3$ then the value of x is :
(a)
(b)
(c)
(d)
$27^{2x - 1} = (243)^3$
$(3^3)^{2x - 1} = (3^5)^3$
$(3)^{3{2x - 1}} = (3)^{5×3}$
3(2x - 1) = 5 × 3
or 2x - 1 = 5 ∴ x = 3
Q-18) The average of marks obtained by 100 candidates in a certain examination is 30. If the average marks of passed candidates is 35 and that of the failed candidates is 10, what is the number of candidates who passed the examination?
(a)
(b)
(c)
(d)
Number of successful students in the exam = x
∴ Number of unsuccessful students = 100 – x
According to the question,
30 = ${35x+10(100-x)}/100$
⇒ 3000 = 35x + 1000 – 10x
⇒ 3000 = 25x + 1000
⇒ 25x = 3000 – 1000 = 2000
⇒ x = $2000/25$ = 80
Q-19) If a man walks 20 km at 5 km/ hr, he will be late by 40 minutes. If he walks at 8 km/hr, how early from the fixed time will he reach?
(a)
(b)
(c)
(d)
Time taken to cover 20 km at the speed of 5km/hr
= 4 hours.
Fixed time = 4 hours - 40 minutes
= 3 hour 20 minutes
Time taken to cover 20 km at the speed of 8 km/hr
= $20/8$ = 2 hours 30 minutes
Required time
= 3 hours 20 minutes - 2 hours 30 minutes
= 50 minutes
Q-20) In how many years will Rs.2,000 amounts to Rs.2,420 at 10% per annum compound interest?
(a)
(b)
(c)
(d)
Using Rule 1,
According to question,
2420 = 2000$(1 + 10/100)^t$
$2420/2000 = (11/10)^t$
or $(11/10)^t = 121/100$
or, $(11/10)^t = (11/10)^2$
t = 2 years
Q-21) A train goes from Ballygunge to Sealdah at an average speed of 20 km/hour and comes back at an average speed of 30 km/hour. The average speed of the train for the whole journey is
(a)
(b)
(c)
(d)
UsingRule 5,
Required average speed
= ${2 × 30 × 20}/{30 + 20}$
[Since, Distance covered is same]
= ${2 × 30 × 20}/50 = 24$ kmph
Q-22) The population of a state increased from 100 million to 169 million in two decades. What is the average increase in population per decade?
(a)
(b)
(c)
(d)
The difference of population in two decades
= 169 – 100 = 69 million
∴ Increase in population in first decade
= $69%/2$ = 34.5%
Q-23) A milkman mixed some water with milk to gain 25% by selling the mixture at the cost price. The ratio of water and milk is respectively
(a)
(b)
(c)
(d)
C.P. of 1 litre of milk= Rs. 100
Mixture sold for Rs. 125
= $125/100 = 5/4$ litre
Quantity of water = $5/4 - 1 = 1/4$ litre
Required ratio = $1/4$ : 1 = 1 : 4
Q-24) The probability that at least one of the events A and B occurs is 0.7 and they occur simultaneously with probability 0.2. Then P($\ov{A}$) + P($\ov{B}$) =
(a)
(b)
(c)
(d)
We have P (A ∪ B) = 0.7 and P (A ∩ B) = 0.2
Now, P(A ∪ B) = P(A) + P(B) - P (A ∩ B)
⇒P(A) + P(B) = 0.9⇒1-P($\ov{A}$) + 1 - P($\ov{B}$) = 0.9
⇒P($\ov{A}$) + P($\ov{B}$) = 1.1
Q-25) If $√3$ = 1.732, then what is the value of ${4 +3√3}/{√{7 + 4 √3}}$ upto three places of decimal ?
(a)
(b)
(c)
(d)
${4 +3√3}/{√{7 + 4 √3}}$
${4 +3√3}/{√{7 + 4 √3}}={4+3√3}/{√{4+3+2×2×√3}}$
=${4+3√3}/{√(2+√3)^2}={4+3√3}/{2+√3}$
${(4+3√3)(2-√3)}/{(2+√3)(2-√3)}$
=$8-4√3+6√3-9$
=$2√3-1=2×1.732-1$
= 3.464 - 1 = 2.464
Q-26) If area of a circle and a square are same, then what is the ratio of their perimeters ?
(a)
(b)
(c)
(d)
Let radius of cirlce is ‘r’ and side of the square is a ∼ then, from question
$π r^2 = a^2 ⇒ r/a = 1/√π$
Now ∼ ${2πr}/{4a} = {2π}/{4√π} = √π/2$
Q-27) The speed of two trains are in the ratio 6 : 7. If the second train runs 364 km in 4 hours, then the speed of first train is
(a)
(b)
(c)
(d)
Using Rule 1,Distance = Speed × TimeSpeed = $\text"Distance"/\text"Time"$ , Time = $\text"Distance"/\text"Speed"$1 m/s = $18/5$ km/h, 1 km/h = $5/18$ m/s
Speed of second train
= $364/4$ = 91 kmph
7x ≡ 91
6x ≡ $91/{7x}$ × 6x ≡ 78 kmph
Q-28) What should come in place of the question mark (?) in the following number series?
8, 15, 28, 53, ?
(a)
(b)
(c)
(d)
(e)
Let x = 8
then 15 = 2x – 1 = y
28 = 2y – 2 = z
53 = 2z – 3 = m
Next term in the pattern should be
2m – 4 = 2 × 53 – 4
= 102
Q-29) A man buys a cycle for Rs.1400 and sells it at a loss of 15%. What is the selling price of the cycle?
(a)
(b)
(c)
(d)
Using Rule 3,If an object is sold on r% Profit.
then,S.P. = C.P$[{100 + \text"Profit%"}/100]$orC.P. = S.P$[100/{100 + \text"Profit%"}]$
Similarly, If an object is sold on r% loss, then
S.P. = C.P.$[{100 - \text"Loss%"}/100]$orC.P. = S.P$[100/{100 - \text"Loss%"}]$
Selling price = $1400 × {100 - 15}/100$
= $1400 × 85/100$ = Rs.1190
Q-30) Ravi buys an article with a discount of 25% on its marked price. He makes a profit of 10% by selling it at Rs.660. The marked price of the article was:
(a)
(b)
(c)
(d)
CP of the article for Ravi
= $660 × 100/110$ = Rs.600
Ravi bought the article at the discount of 25%
75% of marked price = Rs.600
Marked price
= ${600 × 100}/75$ = Rs.800