**SSC2MT51**

- 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**

- 60
^{0} - 30
^{0} **90**- 120
^{0}

**Solution :**

Since QM=MR

hence RQM = 30^{0 } ((180-120)/2=60/2= 30)

= 12/24 = 1/2 =

Hence QM is the angle bisector of angle PQR

Than PQM = RQM = 30^{0}

Than PQR = 60^{0}

So QPR = 180 – (30 + 60)

So QPR = 90^{0}

**SSC2MT53**

- 12
**6**- 8
- 14

**Solution :**

A.T.Q

12×6 M = 18×8 W

72M = 144W

1M = 2W

Total Work = M_{1}D_{1} = 12*6M = 72M

Let us assume together they can finish thrice the work in x days

M_{1}D_{1} = M_{2}D_{2}

72 M × 3 = (16M + 40 W) × x

1M = 2W so 40W = 20M

∴216 = 36x

X = 6 days

**SSC2MT54**

*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 = (BC^{2} + AC^{2} – AB^{2})/(2×BC×AC)

cos C = (144 + 225 – 36)/360 = 333/360 =111/120

sin^{2}C + cos^{2}C = 1

sin^{2}C = 1 – (111/120)^{2} = (120^{2} -111^{2})/120^{2
}=231*9/120^{2
}sin C =

sin C = 0.38

**SSC2MT55**

- 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**

- 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**

- 486 cu. cm.
- 324 cu. cm.
- 144 cu. cm.
- 18 cu. cm.

**Solution : Answer A**

Formulae related to regular tetrahedron:

h=

total area =

volume =

**SSC2MT58**

- 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**

- 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**

**Solution: Answer B**

**Short Trick:**

tanθ = multiple of sinθ/ multiple of cosθ

tanθ = 35/12

Basic Method:

12 cosθ + 35 sinθ = 37

Using formula –

After comparing,

**SSC2MT61**

- 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**

**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**

- 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.

inradius = 7/2

**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

Hence AD= 3.5*3 = 10.5

Now from the right-angled triangle ABD we get,

AB^{2}= AD^{2}+BD^{2}

a^{2}= (21/2)^{2}+(a/2)^{2}

a=

Hence the length of a side is

**SSC2MT64**

- 3:4
- 4:5
**4:3**- 2:3

**Solution :**

From using Formula

Required:

∴ ==

**SSC2MT65**

- 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,

⇒ AD = 21.63 cm

**SSC2MT66**

- 4x
^{2}+ 14 x+ 1=0 **x**^{2}– 14x + 1=0- x
^{2}+ 14x + 1=0 - 4x
^{2}+ 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**

- 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**

- Can not be determined

**Solution: Answer C**

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**

- 800
- 1600
- 3000
**3200**

**Solution :**

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

**SSC2MT70**

- 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

5_{x6} : 8_{x6} (13_{x6 }=78)

1_{x13} : 5_{x13} (6_{x13 }=78)

1_{x26} : 2_{x26} (3_{x26 }=78)

Applying allegation on the quantity of Zinc,

30 13

26

13: 4

**SSC2MT71**

- -∞
- 0
- –2
**–4**

**Solution :**

**Short Method:**

(x-5)(x-9)=0 gives x=5 and x=9

min value =

max value = ∞ (since factor of x^{2} 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.

- 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.

**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.

- 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.

- 50
- 100
- 200
**0**

**Solution :**

Required difference =

(1000 + 800) – (1200 + 600) = 0