Test: CSIR-NET Chemical Sciences Mock Test - 6 - UGC NET MCQ

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, 
(2x+4000) / (3x+4000) = 40 / 57
⇒ 57 × (2x + 4000) = 40 × (3x+4000)
⇒ 6x = 68,000
⇒ 3x = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000

Let the numbers be x and (100-x).
Then,x(100 − x ) = 5 * 495
⇒ x² − 100 x + 2475 = 0
⇒ (x-55) (x-45) = 0
⇒ x = 55 or x = 45
The numbers are 45 and 55 
Required difference = (55-45) = 10

Ratio of initial investments = 1/2 : 1/3 : 1/4 = 6 : 4 : 3.
Let their initial investments be 6x, 2x and 3x respectively.
A : B : C = (6x * 2 + 3x * 10) : (4x * 12) : (3x * 12) = 42 : 48 : 36 = 7 : 8 : 6.
B's share =  Rs. (378 * 8/21) = Rs. 144.

Let sum=Rs.x
C.I. when compounded half yearly = 
C.I. when compounded annually = 

⇒ > x = 20000

Average of 20 numbers = 0. 
Sum of 20 numbers = (0 * 20) = 0. 
It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (- a).

Required number is 29.

Let the total number of children be x.
Then, x⋆(20% of x) = 405
⇒ x∗20×/100 = 405
(1/5)x² = 405
⇒ x = 45
Number of sweets received by each child =20%
of 45 = 9

Given:
Outer circumference: Inner circumference = 11 : 7
Breadth of the path between concentric circles = 20 m
Calculation:
Let outer radius = R
Inner radius = r
Outer circumference : Inner circumference = 11 : 7
⇒ 2 × π × R : 2 × π × r = 11 : 7
⇒ R : r = 11 : 7
The breadth of the path between concentric circles = outer radius - inner radius
⇒ (11 - 7) units = 20 m
⇒ 4 units = 20 
⇒ 1 unit = 5 m
⇒ 7 units = 7 × 5 = 35 m 
∴ The correct answer is 35 m.

The shopkeeper claims to sell 1 kg of pulses, but actually sells only 950 grams.

Let's assume the cost price (CP) of 1 kg of pulses is 100 units of currency.

Since he uses a weight of 950 grams instead of 1000 grams, the effective cost price for 950 grams is:

However, the shopkeeper sells 950 grams for the price of 1 kg, which is 100 units. Thus, the selling price (SP) for 950 grams is 100 units.

Now, we can calculate the gain:

The gain percent is given by:

So, the correct answer is: 5.26%

Concept used:
Probability = Number of students who opted for exactly one subject / Total number of students
Calculation:
Total number of students = 60
Number of students who opted two or more subjects = 40 + 17 + 20 - 60 - 5 = 12
So number of students who opted exactly one subject = 60 - 12 = 48
Required probability = 48/60 = 0.80
∴ The correct answer is 0.80

AC in coverd in 2 hr by 1st train = 72 × 2 = 144 km
BC in coverd by both trains in (11 am to  1: 30pm = 2.5 hr) so speed will be added (opposite direction)
⇒ (72 + 90) × 2.5
⇒ 162 × 2.5 = 405 km
Distance covered after 12 noon = 405 - (72 + 90) = 243 km
Time taken in 243 km (12 pm to 4: 30 pm) = 4.5 hr
Let the reduction be 'y' km/h in each train
So, the speed of both train = 243/4.5 = 54 km
⇒ (72 - y) + (90 - y) = 54
⇒ 162 - 2y = 54
⇒ 2y = 108
⇒ y = 54 km
The speed of the train, after 12 noon, that started from the station A, = 72 - x = 72 - 54 = 18 km/h

Given:
The price of a commodity increased by 5% from 2010 to 2011, 8% from 2011 to 2012 and 77% from 2012 to 2013. 
Formula used:
Average = Sum of observations/Number of observations
Calculation:
According to the question
The average price increases from 2010 to 2013 = (5 + 8 + 77)/3
⇒ (90/3)%
⇒ 30%
∴ The required average is 30%.

Given:
No. of lecturers in state A in 2017 = 10
No. of lecturers in state B in 2019 = 22
Formula used:
Percentage = 
Calculations:
According to the formula,
⇒ Percentage = (10/22) x 100 = 45.45%
⇒ Hence, No. of lecturers in state A in 2017 is 45.45% of state B in 2019. 

Given that A, B, and C are pipes that take a, b, and c hours respectively to fill the pool. According to the problem,
we have: b = a + 20 and b = c + 14
Subtracting these equations, we get:
a + 20 = c + 14 ⇒ a - c = 6 ---- (equation 1)
Also given that A and B together can fill the pool in half the time that is required for C to fill the pool, we can write:
a*b / (a + b) = c / 2
Given b = a + 20, replacing b in above equation we get:
a × (a+20) / (2a + 20) = c / 2
Solving this gives:
a² + 20a = (c + a) ×  c
As a - c = 6 from equation 1, replacing c with (a-6) in above equation we get:
a² + 20a = (a - 6) × a
Upon simplifying, we get:
a² + 20a = a² - 6a
This simplifies to:
26a = 60
Solving further we find:
a = 15 hours
Substituting a = 15 into equations b = a + 20 and b = c + 14 equations, we get:
b = 15 + 20 = 35 hours
c = b - 14 = 35 - 14 = 21 hours
To find out how long it would take for all three to fill the pool together, we add their rates of work. The rate of work is the reciprocal of the time taken.
The combined rate of work = 1/a + 1/b + 1/c = 1/15 + 1/35 + 1/21 = 7/105 + 3/105 + 5/105 = 15/105 = 1/7
So, they will fill the pool together in 1 / (1/7) = 7 hours.

By the time they meet for the second time, both of them, together, have completed exactly 2 rounds, since they are running at different speeds, one of them has completed more than one round and other is yet to complete a round. A runs 900m per meeting. Hence A ran 1800m by the time they met for the second time. B ran a total of 800 m by this time.
Cirumference = 1800 + 800 / 2 = 1300 m

When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine. This reaction involving degradation of amide and is popularly known as Hoffmann bromamide degradation reaction.

Triply bonded compounds cannot exhibit geometrical isomerism as the -C=C- bond in these compounds is linear.
Geometrical isomerism is a type of stereoisomerism in which the molecular formula and structure are same but the relative arrangement of atoms is different. Geometrical isomerism arises commonly in heteroleptic complexes. This type of isomerism arises due to the different possible geometric arrangements for the ligand.

The isotopes of Sulfur and Carbon are given incorrectly.
Isotopes of Carbon : ¹²C₆,¹³C₆,¹⁴C₆
Isotopes of Sulfur: ³²S₁₆,³³S₁₆,³⁴S₁₆,³⁵S₁₆
Atoms of the same element that contains the same number of electrons but a different number of neutrons are termed Isotopes. Thus, isotopes have the same atomic number but different mass number.

macronutrient is a chemical element (e.g. potassium, magnesium, calcium) required in large amounts for plant growth.

The electronegativity of sp² hybridised atom will be 2.75. Fluorine is the most electronegative element.

An optically active substance is one which can rotate the plane of polarisation of plane polarised light. if you shine a beam of polarised monochromatic light (light of only a single frequency – in other words a single colour) through a solution of an optically active substance, when the light emerges, its plane of polarisation is found to have rotated.

The optimum DP value of cellulose is 250. The number of monomeric units contained in the polymer is called degree of polymerisation (DP).

Reduction of nitroalkanes yields amines, as shown in below reaction.

The temperature at which solid and liquid coexist in equilibrium is called melting point of solid or freezing point of liquid.

Secondary air pollutants are poisonous substance formed from primary air pollutants. In bright sun light nitrogen, nitrogen oxides, hydrocarbons and O₂ interact to produce more powerful photochemical oxidants like ozone (O₃), peroxyacetyl nitrate (PAN), aldehydes, sulphuric acid, peroxides, etc.

Pyridine is a basic five membered heterocyclic organic compound with the chemical formula C5H5N. It is structurally related to benzene, with on (=CH–) group replaced by a nitrogen atom.

Rain water is free from hardness and surface impurities. Surface water and sea water is very impure and saline.

Overlapping of sp² orbital of carbon with p-orbital of halogen is one of the reasons. Due to conjugation double bond character in alkyl halide.

ΔH<ΔE only if the number of moles of products is less than the number of moles of the reactants.
If n_P < n_r; Δn_g = n_p−n_r =− ve. where, n_P = No. of moles of products
n_r = No. of moles of reactants, n_g = No. of moles of gas
Therefore, ΔH < ΔE.

As we can see below, carbon attached to the halogen atom is further attached to one carbon atom.
Example: CH₃-X → Methyl halide
CH₃-CH₂-X → Ethyl halide
CH₃-CH₂-CH₂-X → n-Propyl halide.