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-**

- 12(1/2)%
**8(4/7)%**- 14(2/7)%
- Can not be determined

**Solution B:**

**Short Trick:**

**Basic Method:**

^{4/7%}

**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:**

**Basic Method:**

**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:**

*tan B*) = 2

*tan B) = 2*

*1+tan B = 1*

*tan*

*B = 0*

**Basic Solution:**

^{0}

^{0}

**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:**

^{3}) radius 2 3(in the ratio of r) Area 4 9 (in the ratio of the square of r

^{2})

**Basic method:**Let the radius be r

_{1 }and r

_{2}. 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

**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

**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:**

**Basic Method:**

^{nd}year

**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 :**

^{2}= 28

^{2}+ 45

^{2}

**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:**

^{2}= 2- (7/17)

^{2}

^{ =2- 49/289 =529/289 }sin θ+ cos θ= 23/17

**Note: when**

**is asked and**sin θ- cos θ

**is given or vice versa. always remember the sum of their squares would be 2.**

**Basic Method:**

^{2}θ + cos

^{2}θ – 2 sin θ cos θ = 49/289

^{2 }= (sin θ- cos θ)

^{2}+ 2 x 2

^{2 }= 529/289

**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:**

**Basic Method:**

∴Profit % =

**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 :**

**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.

**Basic Method:**

**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.

**Basic Method:**Average age of

**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:**

**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:**

^{= 684/1520×100}

**Basic Method:**

^{100/115}

^{= 684/1520×100}

**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.

**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

^{3}– (y+z-x)

^{3}– (x+z-y)

^{3}– (x+y-z)

^{3 }= (1+1+1)

^{3}– (1+1-1)

^{3}– (1+1-1)

^{3}– (1+1-1)

^{3 }

**SSCMT71- **If x =. Find the value of x.

**20**- 21
- 19
- 18

**Solution:**

**Short Trick:**

**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

**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]

- 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]

**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]

- 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]

- 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%