SSC Quantitative Aptitude Questions #1

0
84

SSC Quantitative Aptitude Questions #1

SSCMT51 – Rohit can do a work in 60 days. Rohit and Kamal can do the same work in 40 days while Vishal and Kamal can do the same in 30 days. Find in how many days Rohit, Kamal, and Vishal can complete the whole work working together.

Options:-

  • 21 days
  • 20 days
  • 22 days
  • 25 days

Solution :

Short Trick:

Let total work = LCM (60,40,30) = 120
efficiencies of respective members
Rohit=2;
Kamal=3-2=1;
Vishal=4-1=3;
Total efficiency=2+1+3=6;
Total Work=120
Time=120/6
=20 Days Answer

Basic Method:

Rohit do his work = 60 days
Rohit’s 1-day work =1/60 (Rohit and Kamal)s 1-day work =1/40
∴ Kamal’s 1-day work =(1/40)-(1/60)
=(3-2)/120
=1/120 (Vishal and Kamal)s 1-day work =1/30 ∴ Vishal’s 1-day work =1/30-1/120 =4-1/120 =3/120
=1/4 Total 1-day work of (Rohit + Kamal + Vishal) =1/60+1/120+1/40 =2+1+3/120
=6/120
=1/20 Number of day’s required to complete the work by them
=1/(1/120) days
=20days Answer


SSCMT52-
The ratio of cost price and the market price of an article is 4:5 and the ratio of % profit and % discount is 5:3, then what is the discount percent?
  • 12(1/2)%
  • 8(4/7)%
  • 14(2/7)%
  • Can not be determined
Solution B:
Short Trick:
Basic Method:
Given- C.P: M.P 4: 5 Profit: Discount 5x: 3x
A.T.Q
4(100+5x)/100 = 5(100-3x)/100(S.P in both cases would be the same)
400 + 20x = 500 – 15x
∴35x = 100
∴x =100/35
=20/7 Required Discount %
= 3× 20/7
=60/7
=84/7%

SSCMT53- If the quantity of pulp in fresh fruit is 20% while in dry fruit it is 80%. Then find the quantity of dry fruit obtained from 140 kg of fresh fruit.

  • 45
  • 55
  • 25
  • 35
Solution :
Short Trick:
20% of F.F. = 80% of D.F. (pulp remains constant)
F.F./D.F.=80/20=4/1
if 4 unit=140Kg then
1 unit=140/4
=35Kg
Note: F.F. means Fresh Fruit and D.F. means Dry Fruit.
Basic Method:
The quantity of pulp in fresh fruit
= 20 % of 140 kg
= 28 kg. A.T.Q.
Quantity of pulp in dry fruit
= 80 %
∵ 80 % = 28 kg 1 %
100% = 28/80X100
=35 Kg.

SSCMT54- Find the value of (A+B) if (1+ tan A)(1+tab B)=2

  • π/2
  • π/4
  • π/6
  • π/3
Solution B:
Short Trick:
put A= 45 and solve (1+1)*(1+tan B) = 2
2 * (1+tan B) = 2
1+tan B = 1
tan
B = 0
B = 0
A+B= 45 + 0
= 45
Basic Solution:
(1+ tan A)(1+ tab B)=2
1+tan A+ tab B+ tan A tan B =2
⇒tan A + tan B = 1- tab A tan B
⇒tan A + tan B / 1-tan A tan B=1
⇒tan (A+B)=1
⇒tan (A+B) = tan 450
⇒(A+B)=450
⇒π/4

 

 

SSCMT55- If the volumes of two sphere are in the ratio of 12: 40.5. The ratio of their surface area is:

  • 4: 9
  • 16: 25
  • 2: 3
  • None of them
Solution:
Short Trick:
Volume 120 405 8 27 (in the ratio of r3) radius 2 3(in the ratio of r) Area 4 9 (in the ratio of the square of r2)
Basic method: Let the radius be rand r2. Required  

 


SSCMT56- If the driver of a train increases the speed of the train from 55km/hr to 70 km/hr, he travels 75 km more in the same time, what is the actual distance traveled by him?

  • 350 k.m
  • 280 k.m
  • 275 k.m
  • 210 k.m
Solution:
Short Trick: since time didn’t change, take it same (1 unit).
Basic Solution: Let the required time be x hr. Then 70*x – 55*x =75 15x = 75 X=5
hr Required 55*5=275 k.m

SSCMT57- If the share of three persons A, B, and C is such that A’s share is 7/11 times of B’s share and C’s share is 8/13 times of B’s share. if the total profit is Rs 6440. What is the share of A?

  • 1940
  • 1820
  • 2020
  • 2180
Solution : Hence A:B:C = 91 : 143 :88 A.T.Q (91+143+88) unit
= 6440 322 unit
= 6440 1 unit
= 20 Required, 91 unit
= 20×91
= Rs.1820.

SSCMT58-If a person invests a certain amount of money in a fixed deposit scheme offering 12% per annum compound interest for the 1st year and 15% per annum for 2nd year and received an amount of Rs.15456 after two years. What is the amount?

  • 14000
  • 11435
  • 11500
  • 12000
Solution :
Short Trick:
Let principle be 100 unit Then Amount
=100+12+16.8
=128.8 unit
128.8 unit = 15456
100 unit
= 15456*100/128.8
=12000 Rs
Basic Method:
Let the principal be x A.T.Q Amount after
the first year + x =1.12x Amount
after 2nd year
=1.288x
∴1.288x =15456 X=
=12000 Rs.


SSCMT59- Find the difference between circumradius and inradius of the triangle whose sides are 28 cm, 45 cm, and 53 cm.

  • 17.5 c.m
  • 16.5 c.m
  • 17 c.m
  • 15.45 c.m
Solution :
Since 532 = 282 + 452
So it is a right angle triangle R (circumradius)
= hypoteneous/2
= 53/2
= 26.5 r (inradius)
=(P+B-H)/2
= (28+45-53)/2
= 20/2 = 10
Required = R–r
= 26.5 – 10
=16.5 cm

SSCMT60-

If sin θ- cos θ = 7/17 (0<Ɵ<90). Then find the value of sin θ + cos θ.

  • 15/17
  • 8/17
  • 23/17
  • 21/15
Solution B:
Short Trick:
(sin θ+ cos θ )2 = 2- (7/17)2 =2- 49/289 =529/289 sin θ+ cos θ= 23/17
Note: when
sin θ+ cos θ is asked and sin θ- cos θ is given or vice versa. always remember the sum of their squares would be 2.
Basic Method:
sin θ- cos θ = 7/17
On squaring both sides
sin2 θ + cos2 θ – 2 sin θ cos θ = 49/289
1-49/289 = 2
sin θ cos θ = 240/289
(sin θ + cos θ)= (sin θ- cos θ)2 + 2 x 2
sin θ cos θ = 49/289 + 2x(240/289) (sin θ + cos θ)2 = 529/289
sin θ + cos θ = 23/17

SSCMT61- When an article is sold for 128 instead of 112, the profit percentage triples. If the same article is sold for 117, then what will be the profit percentage?

  • 8(5/6)%
  • 12(1/2)%
  • 14(2/7)%
  • 16(2/3)%
Solution :
Short Trick:
New S.P = 117
Profit = 117-104 =13
∴ Profit % =13/104 x 100
= 100/8
= 12(1/2)
Basic Method:
Let the profit be x. A.T.Q (112-x) = (128-3x)
∴2x = 16 Then x = 8
So the C.P would be (112-x) = 112-8
= 104
New S.P = 117
Profit = 117-104 =13

∴Profit % =

13/104 x 100
= 100/8
= 12(1/2)

SSCMT62- In the given figure, ∆ABC is inscribed in a circle with O is center and OM is ⏊to AC. If AO= 2cm and AC= 2, then find ∠ABC.

  • 30
  • 60
  • 45
  • 90
Solution :
AM=OM bisects AC into two equal parts) In
⊿AOM,
sinƟ =
∴ Ɵ= 30 And
∠AOC = 2*∠AOM = 60
∠ABC =

 


SSCMT63- A container contains wine and spirit in the ratio of 9: 7. If 6-liter mixture is taken out and the same amount of spirit is poured into the mixture, then the ratio of spirit and wine becomes equal. Find the original quantity of mixture.

  • 64
  • 48
  • 54
  • 50
Solution: Since total mixture remains equal, make them both LCM(16,2)=16 units.
Now new ratio becomes
decrease in quantity of Wine = (9/16) * 6 L
1 unit = 54/16
L; 16
units = 54 L;
Total mixture = 54 L
Basic Method:
Let Initial mixture be 16x+6 L After 6-liter mixture is taken out, the remaining quantity is 16x.
Wine = 9x and Spirit = 7x
Now according to question
9x = 7x + 6
2x = 6x
= 3 Initial solution
= 16x+6
= 16*3+6
= 54

SSCMT64- Two years ago the average age of a family of 8 members was 18 years. After the addition of a baby, the average age of the family is the same today. What is the age of the baby (in years)?

  • 2
  • 1(1/2)
  • 1
  • 2(1/2)
Solution:
Short Trick: Conditions today: People Avg Age 8 20 (without baby) 9 18 (with baby) these 2 years are deducted from everyone’s age due to an addition of the baby. So the baby’s age is 2*8=16 less than the new average.
Baby’s Age
= 18-2*8
=2 years
Basic Method: Average age of
8 members two years ago = 18 Sum of the age of
8 members two years ago = 18×8=144
In these two years, the age of 8 members is also increased. Increase in the age of
8 members in two years = 8×2=16
Sum of the age of 8 members today = 144+16=160
∵After the addition of a baby, the average of the family is 18 again.
∴average age of
9 members today = 18 years Sum of the age of
9 members today = 18×9=162
∴ age of baby = 162-160
=2 years.

SSCMT65- Find the greatest number which when divides 564 and 467 leave remainder 4 and 7 respectively.

  • 30
  • 20
  • 25
  • 35
Solution : 564 – 4 = 560 467 – 7 = 460 We need the greatest such number that divides both 560 and 460. Which is the definition of HCF? HCF (560, 460) = 20 So, the greatest number is 20. Note: LCM is defined as the smallest such number that is completely divisible by 2 or more numbers.

SSCMT66- ABCD is a square and F & E are midpoints of AB and BC then find the Ratio of area of shade part and non-shaded part.

  • 5:3
  • 3:5
  • 10:9
  • 9:10
Solution:
Suppose the length of each line is 4. Hence the total area is 16. Now
area of the shaded
part 1: (1/2)*4*2=4 (right angle triangle) area of shaded
part 2: (1/2)*4*2=4 (right angle triangle) area of shaded
part 3: (1/2)*2*2=2 (right angle triangle)
Shaded: non-shaded (4+4+2) : (16-4-4-2) 10: 6 5 : 3

SSCMT67-What is the remainder when 1553 × 1554 × 1555 is divided by 16?

  • 4
  • 6
  • 12
  • 32
Solution : Remainder when 1553 is divided by 16 = 1 Remainder when 1554 is divided by 16 = 2 Remainder when 1555 is divided by 16 = 3 Then the required remainder = 1x2x3 = 6

SSCMT68- A shopkeeper sells an article at Rs.1748 and gains 15%. if he sells the same article at Rs.2204 find his new profit percentage.

  • 35%
  • 45%
  • 40%
  • 54%
Solution :
Short Trick:
15% = 3/20
CP=20 unit
SP=23 unit
23 units = 1748
20 units = 1748*20/23
CP=1520
New SP= 2204
New Profit % =(2204-1520)/1520×100
= 684/1520×100
=45%
Basic Method:
C.P = 1748x100/115
=1520
New S.P = 2204 Required New profit
=(2204-1520)/1520×100
= 684/1520×100
=45%

SSCMT69- A straight line intersects the x-axis and y-axis at P and Q respectively. If the midpoint of line PQ is (6, 10), then find the area of ∆OPQ.

  • 30
  • 80
  • 120
  • 140
Solution: Let the coordinates of P and Q be (x, 0) and (0, y) respectively.
Now using the midpoint formula we get,
x+0/2 = 6
x =12 and
y+0/2 = 10
y=20
Then the area of ΔOPQ = 1/2 * X * Y
= 1/2 * 12*20
= 120 Sq. unit.

 

SSCMT70- The value of (x+y+z)3 – (y+z-x)3 – (x+z-y)3 – (x+y-z)3 is:

  • 3 XYZ
  • 6 XYZ
  • 12 XYZ
  • 24 XYZ
Solution:
Short Trick: To solve these kinds of questions quickly we can use putting method, put some desirable values in question as well as options such that all options get different numerical values. put
x=y=z=1;
then from question
(x+y+z)3 – (y+z-x)3 – (x+z-y)3 – (x+y-z)= (1+1+1)3 – (1+1-1)3 – (1+1-1)3 – (1+1-1)
= 27-1-1-1=24 from options
24xyz=24

SSCMT71- If x =. Find the value of x.

  • 20
  • 21
  • 19
  • 18
Solution:
Short Trick:
Ans= 20
Note 1:
In the case of ‘+’, the answer is the bigger factor. and in the case of ‘-‘ the answer is the smaller factor. Note 2: the difference between factors should be of 1 exactly. Basic Method:

Squaring both sides but can’t

be negative so x=20

SSCMT72- Direction: Read the following information carefully and answer the questions that follow: The following table shows the number of people in different age groups who traveled by PQR Travels in a month.

[su_table responsive=”yes”]

Number of Travelers
Less than 3 years 85
3-10 years 150
11-45 years 98
40-60 years 127
Above 60 years 53

[/su_table]

What is the total number of passengers below 60 years?
  • 450
  • 460
  • 560
  • 480
Solution: Total number of passengers below 60 years = 85+150+98+127 = 460

SSCMT73- Direction: Read the following information carefully and answer the questions that follow: The following table shows the number of people in different age groups who traveled by PQR Travels in a month.

[su_table responsive=”yes”]

Number of Travelers
Less than 3 years 85
3-10 years 150
11-45 years 98
40-60 years 127
Above 60 years 53

[/su_table]

The average number of travelers of all age groups is:
  • 102.6
  • 98.4
  • 104.2
  • None of these
Solution : Total number of passengers = 513 Average = 513/5 = 102.6

SSCMT74- Direction: Read the following information carefully and answer the questions that follow: The following table shows the number of people in different age groups who traveled by PQR Travels in a month.

[su_table responsive=”yes”]

Number of Travelers
Less than 3 years 85
3-10 years 150
11-45 years 98
40-60 years 127
Above 60 years 53

[/su_table]

How many age groups have less than 130 but more than 90 travelers?
  • 1
  • 2
  • 3
  • 4
Solution: There are two age groups with travelers less than 130 and more than 90. 11-45 years: 98 travelers 46-60 years: 127 travelers

SSCMT75-

Direction: Read the following information carefully and answer the questions that follow: The following table shows the number of people in different age groups who traveled by PQR Travels in a month.

[su_table responsive=”yes”]

Number of Travelers
Less than 3 years 85
3-10 years 150
11-45 years 98
40-60 years 127
Above 60 years 53

[/su_table]

What is the percentage of travelers less than 3 years old out of total travelers?
  • 9.5 %
  • 11.5%
  • 13.0%
  • 16.5%
Solution : Total number of travelers above below 3 years = 85 Total number of travelers 513 Percentage = (85/513) x 100 = 16.5%

LEAVE A REPLY

Please enter your comment!
Please enter your name here