# ssc2mt3

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1333

SSC2MT51

A man spends 30% of his income on house rent. He spends 30% of the remaining income on children’s study and 35% of the whole income on other expenditures. Now he has Rs.12600 left. What is the income of the man?
• 80,000
• 98000
• 90000
• 126000

Solution :

Short Trick:
Let the Income of the man be 100 unit.

Remaining 14 unit = 12600
1 unit = 900
100 unit = 90000

Basic Method:
Let the income of the man be x.
He spends on rent =
He spends on children’s study = (x-=
Other expenditure =
A.T.Q
X- (

∴X= Rs. 90,000

SSC2MT52

In the following figure given a triangle PQR if PM: MR = 1: 2 and QM=MR then find the angle QPR =?
• 600
• 300
• 90
• 1200

Solution :

Since QM=MR
hence RQM = 300  ((180-120)/2=60/2= 30)
= 12/24 = 1/2 =
Hence QM is the angle bisector of angle PQR
Than PQM = RQM = 300
Than PQR = 600
So QPR = 180 – (30 + 60)
So QPR = 900

SSC2MT53

12 men can complete a work in 6 days and 18 women can complete the same work in 8 days. if they work together in how many days 16 men and 40 women can do thrice the work.
• 12
• 6
• 8
• 14

Solution :

A.T.Q
12×6 M = 18×8 W
72M = 144W
1M = 2W
Total Work = M1D1 = 12*6M = 72M
Let us assume together they can finish thrice the work in x days
M1D1 = M2D2
72 M × 3 = (16M + 40 W) × x
1M = 2W so 40W = 20M
∴216 = 36x
X = 6 days

SSC2MT54

The sides of a scalene triangle measure 6cm, 12cm and 15cm. What is the sin of the angle opposite the side measuring 6cm?
• 0.38
• 0.46
• 0.57
• 0.69

Solution :

Let the triangle be ABC with AB = 6cm, BC = 12cm and CA = 15 cm.

Using cosine rule,
cos C = (BC2 + AC2 – AB2)/(2×BC×AC)
cos C = (144 + 225 – 36)/360 = 333/360 =111/120
sin2C + cos2C = 1
sin2C = 1 – (111/120)2 = (1202 -1112)/1202
=231*9/1202
sin C =
sin C = 0.38

SSC2MT55

The average income of P, Q and R are Rs. 14000 per month and the average income of Q, R and S are Rs. 18000 per month. If the salary of S is twice to that of P, then the avg. salary of Q and R is-
• 25000
• 15000
• 19000
• 18000

Solution :

P + Q + R = 14000 X 3
Q + R + S = 18000 X 3
∴S – P = 12000
Since S = 2P
∴2P – P = 12000 ⇒ P = 12000
Required
The avg. salary of (Q + R) is = (42000-12000)/2
= Rs. 15000

SSC2MT56

A shopkeeper purchases an article at a 25% discount and marks its price 40% above its original price. If he sells the article at a discount of 20%, then find the percentage profit of the shopkeeper.
• 46.87%
• 49.33%
• 48.43%
• 50%

Solution :

Short Trick:
Let the Original Price be 100
CP=75
MP=140
SP=112
Profit = 112-75 = 37
Profit % = 37*100/75 =49.33%

Basic Method:
Let the original price be x.
Marked price ==
C.p =
S.P ==
Required
Profit % =

Report Solution

SSC2MT57

If the length of each side of a regular tetrahedron is 18 cm, then the volume of the tetrahedron is.
• 486 cu. cm.
• 324 cu. cm.
• 144 cu. cm.
• 18 cu. cm.

Formulae related to regular tetrahedron:
h=
total area =
volume =

SSC2MT58

N is the greatest number that will divide 734, 761 and 1463 leaving the same remainder in each case. What is sum of digits of N?
• 10
• 7
• 8
• 9

Solution :

Value of N = H.C.F of (761-734), (1463-761) and (1463-734)
= H.C.F of 27, 702, 729 = 27
Sum of digits = 2+7 = 9

SSC2MT59

The population of a city is 8000 in 2015. The male population grew by 8% annually and the female population increased by 12% annually. If the number of males in 2016 is 5940, what is the combined rate of growth of population from 2015 to 2016?
• 10%
• 9.75%
• 9.50%
• 9.25%

Solution :

After 8% increase, the male population is 5940.
Or, 1.08 × male population in 2015 = 5940
Male population in 2015 = 5500
Female population in 2015 = 8000 – 5500 = 2500
Female population in 2016 = 1.12 × 2500 = 2800
Total population in 2016 = 5940 + 2800 = 8740
Percentage growth = [(8740 – 8000)/8000] × 100 = 9.25%

SSC2MT60

If 12 Cosθ + 35 Sinθ = 37 then find the value of tanθ.

Short Trick:
tanθ = multiple of sinθ/ multiple of cosθ
tanθ = 35/12

Basic Method:

12 cosθ + 35 sinθ = 37

Using formula –

After comparing,

SSC2MT61

If the savings of Richa and Abhishek per month are in the ratio of 4:7. If by the end of the financial year, savings of each of them is increased by Rs. 10,000, then the ratio becomes 23:27. What is Richa’s present saving?
• Rs. 12763
• Rs.12246
• Rs. 12247
• Rs. 13018

Solution :

⇒ Let the original saving of Richa and Abhishek be 4x and 7x.
⇒ According to the question given in the problem,

⇒ 108x + 270000 = 161x + 230000
⇒ 53x = 40000
⇒ x= 754.71
∴ Richa’s present saving = 4x + 10000
= 3018 + 10000
= Rs. 13018

SSC2MT62

If the total cost price of a pen and a pencil is 500 RS. There is a loss of 10% on a pencil while the shopkeeper makes a profit of 20% on the pen. If there is an overall profit of 40 RS. Find the cost price of the pen.
• 300 RS
• 250 RS
• 400 RS
• 340 RS

Solution :

Short Trick:

CP of pen = 300

Basic Method:
Let CP of Pen = x and CP of Pencil = y

2x – y = 400 (equation 1)
And given x + y = 500 (equation 2)
By adding equation 1 and equation 2
⇒3x = 900
X= 300 RS

SSC2MT63

One of the sides of an equilateral triangle is the line 6x+8y+7=0 and its centroid is at P(2,2). Find the length of a side.
• 6
• 6
• 7
• 7

Solution :

Short Trick:
length of the perpendicular from P(2,2) upon the line 6x+8y+7=0,
=== 7/2
Since in the case of equilateral triangle centroid is same as incentre.

Basic Method:
Let the equation of the side BC of the equilateral triangle ABC be 6x+8y+7=0
If AD ⊥ BC, then BD= ½ BC= a/2 ……………(let a be the length of a side of the equilateral triangle)
Now for an equilateral triangle, we know that P lies on the AD and PD= 1/3 AD
Now, length of the perpendicular from P(2,2) upon the line,
=== 3.5
Now from the right-angled triangle ABD we get,
a2= (21/2)2+(a/2)2
a=
Hence the length of a side is

SSC2MT64

Two trains started Simultaneously at the same time, one from A to B and other from B to A. If they arrived at B and A respectively in 36 hrs and 64 hrs after passing each other, the ratio of the speed of two trains is.
• 3:4
• 4:5
• 4:3
• 2:3

Solution :

From using Formula

Required:
==

SSC2MT65

The radii of two concentric circles are 15 cm and 9 cm. AB is the diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D & the bigger circle at E. Point A is joined to D. The length of an AD is
• 27.63 cm
• 25.50 cm
• 22.47 cm
• 21.63 cm

Solution :

⇒ OA = OB = 15cm
⇒ OD = 9 cm, AE = 18 cm (from similarity of triangles)
⇒ In ΔODB

⇒ BD = DE = 12 cm
⇒ In ΔAED,

SSC2MT66

Find the quadratic equation whose roots areand.
• 4x2 + 14 x+ 1=0
• x2 – 14x + 1=0
• x2 + 14x + 1=0
• 4x2 + 14x + 2=0

Solution :

A.T.Q,
Sum of roots = 7 +4
Product of roots = (7+4)(7-4=49-48 =1
Required equation =
=+ 1=0

SSC2MT67

The speed of a boat in still water is 20% less than its speed downstream. If it covers 20 km in upstream in 1 hr then what is the speed of the current?
• 4.5

Solution: Anser B

Let the speed of the boat in still water be x km/h and speed of current be y km/h
According to the question

4x + 4y = 5x
x = 4y
Speed in upstream = 20 km/h
x – y = 20
4y – y = 20
3y = 20

y =km/h

SSC2MT68

Three containers P, Q and R which contain spirit, water, and alcohol respectively. the quantity of each is 140 ltr. If 10% spirit is taken out and poured into container Q. Then, again 10% from Q is transferred to R, from which again 10% is transferred to P. What is the proportion of water in container P at the end of the process?
• Can not be determined

Spirit Reamining after 10% of the content of P taken out = 126 ltr
Now spirit and water in Q = 140 ltr water and 14 ltr spirit
10% of Content of Q = 14 Ltr water and 1.4 ltr spirit
After 10% of content Q poured into content R, then the content of R = 140 alcohol,14 ltr water,1.4 ltr spirit
10% of content R includes 14 ltr alcohol,1.4 ltr water, 0.14 spirit
Now after pouring 10% of the content of R into P, the content of P = (126+0.14) ltr spirit, 14 ltr alcohol and 1.4 ltr water
Proportion of water in container P ===

SSC2MT69

Amount of a sum of money at the yearly rate of compound interest after two years is Rs 3528 and amount after three years of the same amount at the same rate is 3704.4 then what is the original amount?
• 800
• 1600
• 3000
• 3200

Solution :

Interest for 1 year on 3528 = 3704.4 – 3528 =176.4

SSC2MT70

An alloy A contains zinc and brass in ratio 5:8. And alloy B contains zinc and brass in 1:5. In what ratio alloy A and alloy B is mixed so that the mixture contains zinc and brass in 1:2.
• 13:5
• 13:4
• 4:13
• 5:13

Solution :

The ratio of zinc and brass
5: 8 (13)
1: 5 (6)
1: 2 (3)
Taking LCM of 13, 6 and 3 = 78
Now let us try to make all of them 78, to avoid fractions during the calculations.
Ratio of zinc and brass
5x6 : 8x6 (13x6 =78)
1x13 : 5x13 (6x13 =78)
1x26 : 2x26 (3x26 =78)

Applying allegation on the quantity of Zinc,
30 13
26
13: 4

SSC2MT71

Find the minimum value of given equation.
• -∞
• 0
• –2
• –4

Solution :

Short Method:
(x-5)(x-9)=0 gives x=5 and x=9
min value =
max value = ∞ (since factor of x2 is positive)

Basic Method:

After comparing this equation from getting

min value of equation

SSC2MT72

Direction: In the line graph data is given for the number of passed, failed, and absent students. Study the data carefully and answer the related questions.
What is the passing percentage of students from all the schools together?
• 43.4%
• 42.5%
• 47.3%
• 44.3%

Solution :

Required % =

== 44.3%

SSC2MT73

Direction: In the line graph data is given for the number of passed, failed, and absent students. Study the data carefully and answer the related questions.
What is the difference between the average number of passed and failed students in all schools together?
• 20
• 50
• 100
• 200

Solution :

Required difference =

=

= 20

SSC2MT74

Direction:In the line graph data is given about the number of passed, failed, and absent students. Study the data carefully and answer the related questions.
Number of failed students is what percent of appeared students in A ?
• 41.5%
• 42.5%
• 45.5%
• 55.5%

Solution :

Required % =
= 55.5%

SSC2MT75

Direction:In the line graph data is given about the number of passed, failed, and absent students. Study the data carefully and answer the related questions.
What is difference between no of appeared students in A and B ?
• 50
• 100
• 200
• 0

Solution :

Required difference =
(1000 + 800) – (1200 + 600) = 0