MAT Mock Test - 10 - MAT MCQ

The use of the word 'babble (sound word)' in the blank space of the sentence is appropriate.

The babble of strange and unfamiliar dialects at the market held my interest.

Babble: talk rapidly and continuously in a foolish, excited, or incomprehensible way.

Let's look at the meaning of the given idiom:

A Hard nut to crack - a problem that is very difficult to solve or a person who is very difficult to understand. For Example: The test problem was a hard nut to crack.

Alex Murphy is converted into a Robocop.

According to the line of the passage, Morton has Murphy's corpse converted into RoboCop, turning him into a powerful and heavily-armored cyborg with no memory of his former life.

By the above-mentioned bold line, it is clear that Alex Murphy is transformed into Robocop after his death.

Take in is a correct antonym of the word supplant.

Take in means allow someone to stay in a place. Example- Our club plan to take in 20 new members.

Hence, the correct option is (B).

It is mentioned in the second line of the first paragraph that "Older and wiser and better people (his aunt), had told him that there could not possibly be a frog in his bread-and-milk."

Option (A), (C) and (D) are incorrect as it does not provide us with the most appropriate answer.

The first statement is B as mentioned and it introduces us to the topic of the passage which is a particular personality. F follows B as it talks about her appointment. C follows F as it mentions a reaction of a person to her appointment. E is the fourth statement as mentioned in the passage. D is the fifth statement as it talks on the same note as that of E. A is the sixth statement and G follows A as the words ‘in return’ are mentioned here.

Thus the correct order is BFCEDAG.

The SECOND statement is F.

It is clearly mentioned in the third sentence of the third paragraph of the passage that 'The danger comes only from within; it is a danger from among us.'
Thus, it is clear that the greatest disadvantage of the present time is the danger that comes from within.
It is so because the danger that is outside can be tracked and handled but the danger from within is unprecedented.

Option (C) is false according to the passage because the correct statement would be that most people share their money unequally.
Let's refer to the last paragraph of the passage -  "Today, the sharing-out of money – the sharing-out of food, clothing, houses, and books, is still very unfair. In England alone, in sharing out of the National Income, we find that one half is divided among every sixteen people and the seventeenth person gets the other half. So while some few people live in luxury, many have not even enough to eat and drink and wear. Again, in England today, thousands of people live in dreadful surroundings. There are many families of five or six persons who live in a single room; in this same room, they are born, and in this same room they die, because, they are too poor to afford another room. Until everyone gets his proper share of necessary and delightful things, our civilization will not be perfect."

Order of scoring marks:

1) Meeta's marks are less than Sita's.

Meeta Sita Geeta...(i)

2) Geeta's marks are more than Sita but less than Seema.

Geeta Seema...(ii)

3) Veena has less marks than Meeta.

Veena Mita...(iii)

From the above three equations:

Veena Meeta Sita Geeta Seema

So, the third place belongs to Sita.

From this information, we cannot assume that the next generation will have renewable source of energy as there is no inclusion of renewable energy in the given statement.

Therefore, Assumption I is not implicit.

On the other hand, We can assume that if non - renewable sources of energy go exhausted, the next generation will not be able to use them and will have to think of alternatives as non - renewable energy will extinct if we are not saving it.

Therefore, Assumption II is implicit.

As per the English alphabetical series and its place value.

The logic followed here is:

D + 2 = F; F + 2 = H; H + 2 = J; J + 2 = L

A + 2 = C; C + 2 = E; E + 2 = G; G + 2 = I

G + 2 = I; I + 2 = K; K + 2 = M; M + 2 = O

So, “HEK” is the correct answer.

As,

ROK = R + O + K = 18 + 15 + 11 = 44

and

MIG = M + I + G = 13 + 9 + 7 = 29

Similarly,

TAL = T + A + L = 20 + 1 + 12 = 33

Given, 
A’s rank from left end = 17th
A’s rank from right end = 18 + 7 = 25th
Total number = A’s rank from left end + A’s rank from right end -1
Total number = 17 + 25 - 1 = 41

Quantity I:

Sonia’s share = Rs. [5000 x (1/2)] = Rs. 2500

Amit’s share = Rs. [5000 x (1/2)] = Rs. 2500

Twice the share of Sonia = Rs. 5000

⇒ Difference between twice the share of Sonia and the share of Amit = Rs. 5000 – Rs. 2500 = Rs. 2500

Quantity II:

Let the Y’s share be Rs. x

X’s share + Y’s share + Z’s share + A’s share = Rs. 1,21,000

⇒ (1 × x/2) + x + 50,000 + (x/2) = Rs. 1,21,000

⇒ (x/2) + x + 50,000 + (x/2) = Rs. 1,21,000

⇒ x + x = Rs. (1,21,000 – 50,000)

⇒ 2x = Rs. 71,000

⇒ x = Rs. 35,500

∴ Quantity I < Quantity II

Given,
Ram: Sita = 2 : 5
2 Men +5 Women = 10 days
Let the efficiency of a man = 2x
Then,
The efficiency of a women = 5x
⇒ When a group of 2 men and 5 women (Equivalent to sita)
together can complete a piece of work in 10 days
⇒ Total units of work done in 10 days = (2 × 2x × 10) + (5 × 5x × 10) 
⇒ Total units of work done in 10 days = 290x units
Quantity I:
⇒ The efficiency of a group of a Ram and three women equivalent
to Sita = 2x + (3 × 5x)
⇒ Efficiency = 17x
⇒ required number of days = 290x/17x
⇒ required number of days ≈17 days
∴ The required number of days is 17 days.

Quantity II:

If the efficiency of a Ram was increased by 30% and the efficiency
of a Sita by 40%, then
The new efficiency of a Ram = 130% of 4x = 5.2x
And the new efficiency of a Sita = 140% of 3x = 4.2x 
⇒ The new efficiency of Ram and Sita = 5.2x + 4.2x 
⇒ The new efficiency of Ram and Sita = 9.4x 
 ⇒ 75% of the work = 75%  of 140x 
= 105x units
The required number of days = 105x/9.4x 
⇒ Required number of day ≈11 days
∴ Required number of days is 11 days.
∴ Quantity I >Quantity II

From statement I:
The selling price of the milk = 2 × Cost price of milk Profit

From statement II:
Let the C.P. of pure milk per liter be x.
8 liters of mixture = 2 liters of water +6 liters of milk
Cost price of 8 liters of mixture = C.P. of 2 liter water + C.P. of 6 liter milk = 0 + 6x = 6x
From I, S.P. of 8 liter of mixture = S.P. of 8 liters of pure milk  = 8 × 2x = 16x

Profit%
∴The data in either statement I alone or in statement II alone are sufficient to answer the question.

Quantity A:

We know, T_n = a + (n - 1)d

S_n = n/2 × (2a + (n - 1)d)

Where, a = first term and d = common difference

[a + (25 - 1)d] - [a + (18 - 1)d] = 28

⇒ d = 4

[a + (30 - 1)4] + [a + (6 - 1)4] = 158

⇒ a = 11

S_n = 2070

⇒ (2070 = n/2) × [2 × 11 + (n - 1)4]

⇒ 4140 = 22n + 4n² - 4n

⇒ 2n² + 9n - 2070 = 0

⇒ 2n² - 60n + 69n - 2070 = 0

⇒ 2n (n - 30) + 69 (n - 30) = 0

⇒ n = 30

Number of terms in series when sum of terms of series is 2070 = 30.

Quantity B:

We know,

Surface area of sphere = 4 × 22/7 × radius²

Volume of cylinder = 22/7 × radius² × height

Volume of cube = side³

⇒ 154 = 4 × 22/7 × radius²

⇒ Radius = 3.5 cm

⇒ Radius of cylinder = 3.5 cm

Volume of cylinder = 22/7 × 3.5 × 3.5 × 12 = 462 cubic cm

Volume of small cube = 1 cubic cm

⇒ Number of small cubes formed after melting large cylinder = 462/1 = 462 

∴ Quantity A < Quantity B.

Marked price of product P = Rs.1200 / kg
He promises to sell product P at 25% loss.
⇒ Selling price = Cost price -25% of cost price
⇒ 1200 = 0.75 x Cost price
⇒ Cost price of product P will be = 1200/0.75 = Rs. 1600 / kg
He cheated by 40% while selling the product.
⇒ Actual quantity given for Rs. 1200 =
1000 - 40% of 1000 = 0.6 x 1000 = 600gm
⇒ Cost price of 600gm of product P = (1600/1000) x 600 = Rs. 960
⇒ Percentage profit earned by selling product P 
=  x 100 = 25%
Marked price of product Q = Rs. 810 / kg
He promises to sell product Q at 10% loss.
⇒ Cost price of product Q will be = 810/0.9 = Rs. 900 / kg
He cheated by 20% while buying the product Q.
Actual quantity bought for Rs. 900 = 1000 + 20% of 1000 = 1000 x 1.2 = 1200 gm
He cheated by 10% while selling the product Q.
⇒ Actual quantity given for Rs. 810 = 1000 - 10% of 1000 = 1000 x 0.9 = 900 gm
⇒ Cost price of 900 gm of product Q =
(900/1200) x 900 = Rs. 675
⇒ Percentage profit earned by selling product Q =  x 100 = 20%
⇒ Required difference = 25 - 20 = 5%
∴ Difference in profit percentage while selling product P and Q is 5%.

We know, 37.5% = (3/8)
Let Cost price be 8p
⇒ Marked price = Cost price + (3/8) x cost price
= (11/8) x 8p = 11p
⇒ 11p = 1650
⇒ p = Rs. 150
⇒ Cost price = 150 x 8 = Rs. 1200 / kg
He is cheating by 20% while buying product T.
⇒ Actual quantity bought for Rs. 1200
= 1000 + 20% of 1000 = 1.2 x 1000 = 1200 gm
He is cheating by 10% while selling the product T.
⇒ Actual quantity sold for Rs. 1650 =
1000 - 10% of 1000 = 0.9 x 1000 = 900 gm
⇒ Cost price of 900 grams of product T
= (1200/1200) x 900 = Rs. 900
⇒ Profit percentage
=  x 100 = 250/3% = %
∴ x = % and cost price per kg is Rs. 1200

Since diagonal of a parallelogram bisect each other, 

∴ BO = OD = 18/2 = 9 cm

P = centroid of ΔADC

OP = 1/3 OD = 1/3 X 9 = 3 cm

Q centroid of ΔABC

OQ = 1/3 OB = 1/3 X 9 = 3 cm

∴ PQ = OP + OQ = 3 + 3 = 6 cm

Let the sum of money P = Rs. A

Let the rate of interest R = r %

Case 1:

Time T₁ = 5 years

Simple Interest 

Case 2:
Time T₂ = 20 years

Simple Interest 

Required ratio = SI₁ : SI₂ 

= 1 : 4

Let AF = FB = a and BD = DC = b

In ΔABD, AD² = AB² + BD² ... (1)

⇒ AD² = 4a² + b²

In ΔBFC, FC² = FB² + BC²

⇒ FC² = a² +4 b² ...(2)

Adding equation 1 and 2 ,

⇒ AD² + FC² = 5a² +5 b²

Also, AC² = 4a² + 4b² 

⇒ 4 (AD² + FC²) = 5AC²

Given,
ΔABC is a right angled triangle 90° at B.
AB = 8 cm
BC = 15 cm (Hypotenuse)² = (Perpendicular)² + (Base)² 
Using Pythagoras theorem
(AC)² = (AB)² + (BC)² 
⇒ (AC)² = 8² + 15² 
⇒ (AC)² = 64 + 225
⇒ (AC)² = 289
AC = 17
From the following figure,

P, Q and R are three points where circles meet.
Le AP be r cm, then
⇒ BP = BQ = (8 – r)
⇒ QC = 15 – (8 – r)
⇒ QC = 15 – 8 + r
⇒ QC = 7 + r
⇒ QC = RC = 7 + r
⇒ AR = 17 – (7 + r)
⇒ AR = 17 – 7 – r
⇒ AR = 10 – r
⇒ AR = AP
⇒ r = 10 – r
⇒ 2r = 10
⇒ r = 5 cm
Radius of circle of centre A = 5 cm
Radius of circle of centre B = 8 – 5 = 3 cm
Radius of circle of centre C = 7 + 3 = 10 cm
∴ Radius of smallest circle is 3 cm.

Given:

Now

Put

Putting value of n and m in the equation

Given:
Amount, A = Rs. 7000
Time, T = 4 years
Rate, T = 10%
Formula used:
Simple Interest
Amount = Principal + Simple Interest
Let the sum of money be Rs. P.
Amount = Principal + Simple Interest

⇒ 7000 = 7P/5
⇒ 1000 x 5 = P
⇒ P = 5000
∴ The sum of money is Rs. 5000.

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• At higher gradients, downward, vertical erosion is more dominant. This produces V-shaped valleys
• The V-shaped valley is typical of one that has been carved by flowing water.
• The erosion is more pronounced when the water flow is a heavy one, and the water carries suspended particles (sedimentary load).

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